Download 5 Congruence of Segments, Angles and Triangles

5
Congruence of Segments, Angles and Triangles
This section deals with David Hilbert’s axiomatization of neutral geometry, and follows
quite closely Hilbert’s foundations of geometry.
5.1
Congruence of segments
Proposition 5.1 (Congruence is an equivalence relation). Congruence is an equivalence relation on the class of line segments.
Question. Which three properties do we need to check for a congruence relation?
Answer. For an arbitrary relation to be a congruence relation, we need to check
(a) reflexivity: Each segment is congruent to itself, written AB ∼
= AB.
(b) symmetry: If AB ∼
= AB.
= A B , then A B ∼
(c) transitivity: If AB ∼
= CD, and CD ∼
= EF , then AB ∼
= EF .
Proof of reflexivity, for a bottle of wine from Hilbert personally. 23 We need to show that
each segment is congruent to itself. Take any given segment AB. We transfer the segment to a ray starting from any point C. By Hilbert’s axiom III.1, there exists a point
D on that ray such that AB ∼
= CD. Now use Hilbert’s axiom III.2:
∼
∼
"If A B = AB and A B = AB, then A B ∼
= A B ."
Hence, for a case with other notation, AB ∼
= CD and AB ∼
= CD imply AB ∼
= AB.
Proof of symmetry. Assume that AB ∼
= CD. Because of reflexivity CD ∼
= CD. Once
more, we use Hilbert’s axiom III.2:
"If A B ∼
= AB and A B ∼
= AB, then A B ∼
= A B ."
Hence CD ∼
= CD and AB ∼
= CD imply CD ∼
= AB.
Proof of transitivity. Assume AB ∼
= CD and CD ∼
= EF . Because of symmetry EF ∼
=
CD. Now we use Hilbert’s axiom III.2:
"If A B ∼
= AB and A B ∼
= AB, then A B ∼
= A B ."
For a case with other notation, this means that AB ∼
= CD and EF ∼
= CD imply
∼
AB = EF , as to be shown.
23
which one can take gracefully as a thank-you—for translating from page 15 of the millenium edition
of ”Grundlagen der Geometrie”–etc.
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Proposition 5.2 (Existence and uniqueness of segment transfer). Given a segment AB and given a ray r originating at point A , there exists a unique point B on
the ray r such that AB ∼
= A B .
Question. Which part of this statement is among Hilbert’s axioms? Which axiom is
used?
Answer. The existence of the segment A B is postulated in Hilbert’s axiom of congruence III.1.
Question. How does the uniqueness of segment transfer follow? Which axioms are
needed for that part?
Answer. The uniqueness of segment transfer follows from the uniqueness of angle transfer, stated in III.4, and the SAS Axiom III.5.
Figure 5.1: Uniqueness of segment transfer
Proof of uniqueness. Assume the segment AB can be transferred to the ray r from A
in two ways, such that both AB ∼
= A B and AB ∼
= A B . We choose a point C not on
the line A B . One obtains the congruences
A B ∼
= A B , A C ∼
= A C , ∠B A C ∼
= ∠B A C We are using axiom (III.2), the fact that a segment is congruent to itself, and an angle is
congruent to itself by the last statement of axiom (III.4). By the axiom axiom III.5 for
SAS, this implies ∠A C B ∼
= ∠A C B . By the uniqueness of angle transfer, as stated
−−→ −−−→
in axiom III.4, this implies that the rays C B = C B are equal. Hence B = B is the
−−→
−−→
unique intersection point of the two different rays r = A B and C B . We have shown
that AB ∼
= A B and AB ∼
= A B imply B = B . Thus segment transfer is unique.
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Proposition 5.3 (Subtraction of segments). Given are three points on a line such
that A ∗ B ∗ C, and two points B and C on a ray emanating from A . Suppose that
AB ∼
= A C = A B , AC ∼
then BC ∼
= B C and A ∗ B ∗ C follow.
Figure 5.2: Segment subtraction
−−→
Proof. On the ray originating from B , opposite to B A , we transfer segment BC. We
get the seventh point S such that BC ∼
= B S, and A ∗ B ∗ S. By Hilbert’s axiom III.3
on segment addition, AB ∼
= B S imply now AC ∼
= A S.
= A B and BC ∼
On the other hand, it is assumed that AC ∼
= A C . At first, it seems that the location
of C could be like C ? or C ? in the drawing. But notice, as explained in problem 0.2:
the transfer of segment AC is unique. Hence we get C = S. Finally BC ∼
= B S and
A ∗ B ∗ S imply BC ∼
= B C and A ∗ B ∗ C , as to be shown.
Definition 5.1 (Segment comparison). For any two given segments AB and CD
we say that AB is less than CD, iff there exists a point E between C and D such that
AB ∼
= CE and C ∗ E ∗ D. In this case, also say that CD is greater than AB. We write
CD > AB or AB < CD. equivalently.
Proposition 5.4 (Segment comparison holds for congruence classes). Assuming
AB ∼
= A B and CD ∼
= C D , we get: AB < CD if and only if A B < C D .
Figure 5.3: Segment comparison for congruence classes
−−→
Proof. Transfer segment AB onto ray CD, and get AB ∼
= CE. The assumption AB <
−−→
CD implies that C ∗ E ∗ D. Too, we transfer segment A B onto ray C D , and get
A B ∼
= C E .
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We now use segment subtraction for points C, E, D and C , E , D . Since CD ∼
= C D
by assumption, and CE ∼
= C E —by assumption and construction and
= AB ∼
= AB ∼
transitivity—, and C ∗ E ∗ D; segment subtraction yields ED ∼
= E D and C ∗ E ∗ D .
Thus E lies between C and D , from which we conclude A B < E D .
Proposition 5.5 (Transitivity of segment comparison). If AB < CD and CD < EF ,
then AB < EF .
−→
Proof. I can assume that all three segments lie on the same ray AB, and A = C = E.
(Some transferring of segments will produce segments congruent with the given ones
which satisfy these requirements.) Now, having done this, we get AB < AD and
AD < AF . By definition this means A ∗ B ∗ D and A ∗ D ∗ F .
Question. Express these order relations in words.
Answer. Point B lies between A and D, and point D lies between A and F .
By Theorem 5 from Hilbert’s foundations, any four points on a line can be notated
in a way that all four alphabetic order relations hold. (I shall that statement the ”Four
point Theorem”). Now the four points A, B, D, F satisfy the order relations A ∗ B ∗ D
and A ∗ D ∗ F , which implies they are already notated in alphabetic order. Hence the
two other order relations follow: A ∗ B ∗ F and B ∗ D ∗ F . But A ∗ B ∗ F means by
definition that AB < AF , as to be shown.
Figure 5.4: Transitivity of segment comparison
Corollary 4. If AB < AD and AD < AF , then BD < BF .
Proposition 5.6 (Any two segments are comparable). For any two segments AB
and CD, one and only one of the three cases (i)(ii)(iii) occurs:
Either (i) AB < CD
or (ii) AB ∼
or (iii) CD < AB.
= CD
−−→
Proof. Transfer segment AB onto the ray CD. We get AB ∼
= CB . Now the two
segments CB and CD start at the same vertex and lie on the same ray.
By Theorem 4 from Hilbert’s foundations, for any given three points on a line, exactly
one lies between the other two. Hence the three points C, B , D can satisfy
either (i) C ∗ B ∗ D.
or (ii) B = D.
or (iii) C ∗ D ∗ B . These cases
correspond to the three cases as claimed. Indeed, in case (i), C ∗ B ∗ D implies by
definition AB < CD. Indeed, in case (ii), B = D implies AB ∼
= CB = CD by
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Figure 5.5: All segments are comparable
construction. An explanation may be needed in case (iii): One transfers segment CD
−→
back onto the ray AB and gets CD ∼
= AB, we can now use segment
= AD0 . Since CB ∼
∼
subtraction, and get DB = D0 B and A ∗ D0 ∗ B. Because of the last order relation,
−→
transferring segment CD back to ray AB confirms CD < AB.
Figure 5.6: To extend a line—.
−−→
Lemma 5.1 (”simple fact”). Given a segment AB and a ray CD, there exists a point
R on this ray such that the segment DR is longer than the segment AB.
Proof. By the axiom of order (II.2), there exists a point P such that C ∗ D ∗ P . By the
axiom of congruence (III.1), there exists a point Q on the line CD, lying on the same
side of D as point P , such that AB ∼
= DQ. By the axiom of order (II.2), there exists a
point R such that P ∗ Q ∗ R.
−−→
In the drawing on page 136, the given ray CD is extended to obtain a point P outside
−−→
the segment CD. Next we transfer the given segment AB onto the extension ray DP .
−→
Finally, the ray P Q can be extended even more to get a point R outside the segment
P Q.
The construction has produced points C, D, Q, R on a line satisfying the first two
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order relations below. We invoke the Four point Theorem, Proposition 3.3.
C∗D∗Q
and
D∗Q∗R imply
C∗D∗ R
−−→
The third order relation confirms that point R lies on the ray CD. The order relation
D ∗ Q ∗ R means that the segment DR is longer than the segment DQ. By construction,
the segments AB ∼
= DQ are congruent. We see that
DR > DQ ∼
= AB
By Proposition 5.4, segment comparison holds for congruence classes. We conclude
DR > AB, as required.
Remark. What does Euclid’s Second Postulate ”To extend a line” really mean?
A bid of thought shows that this postulate means more than just Hilbert’s axiom
of order (II.2). Remember Achilles and the tortoise! Indeed, we often need to extend
a given line beyond a given segment. This process involves a combination of the axiom
of order (II.2) with the axiom of congruence (III.1) for the segment transfer.
Another situation occurs when one knows, for any other reason, some point E beyond
which one has to extend the line. Even in that situation, we need to invoke the Fourpoint Theorem (Hilbert’s Proposition 5) for the possibility of further extension beyond
E. So we end up with different possible interpretations of Euclid’s postulate!
Definition 5.2 (Sum of segments). The sum of any two segments AB and EF is
defined to be the segment AC were C is the point such that A ∗ B ∗ C and EF ∼
= BC.
−→
In other words, the sum segment AB + EF is obtained by extending the ray AB to
the point C by a segment BC congruent to the second summand EF .
Proposition 5.7 (Congruence classes of segments are an ordered Abelian
group). The sum of segments is defined on equivalence classes of congruent segments.
These equivalence classes have the following properties:
Commutativity a + b = b + a.
Associativity (a + b) + c = a + (b + c)
Comparison Any two segments a and c satisfy either a < c, or a = c or a > c.
Difference a < c if and only if there exists b such that a + b = c.
Comparison of Sums If a < b and c < d, then a + c < b + d.
Hence theses equivalence classes are an ordered Abelian group.
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Proof. By Hilbert’s axiom of congruence III.3, the sum is defined on equivalence classes
of congruent segments. Now we check the items stated:
Commutativity: Let segment AB represent the equivalence class a. By axiom III.1,
we can choose point C such that A ∗ B ∗ C, and segment BC represents the
equivalence class b. Since BC = CB, and BA = AB, these two segments represent
the classes b and a. By the definition of segment addition CA = CB + BA, and
the segment represents b + a. Hence AC = CA implies that a + b = b + a.
Associativity: We construct (a + b) + c. To this end, choose a segment AB of congruence class a, in short AB ∈ a. Again, choose point C such that A ∗ B ∗ C, and
segment BC ∈ b. Furthermore, choose point D such that A ∗ C ∗ D, and segment
CD ∈ c. We now know that
AD ∈ (a + b) + c
On the other hand, we begin by constructing at first b + c. By the four-point
theorem, A ∗ B ∗ C and A ∗ C ∗ D imply B ∗ C ∗ D. Hence BD ∈ b + c. The
construction of a + (b + c) is finished by finding point H such that A ∗ B ∗ H, and
segment BH ∈ b + c. We now know that
AH ∈ a + (b + c)
The uniqueness of segment transfer implies H = D. Finally, we see that AD =
AH, and hence (a + b) + c = a + (b + c).
Comparison: Let any segments a and c be given. These equivalence classes can be
−→ −→
represented by segments AB ∈ a and AC ∈ c on the same ray AB = AC. The
three point theorem implies that either B = C, or A ∗ B ∗ C, or A ∗ C ∗ B. Hence
either a = c, or a < c, or c < a. In the last case a > c as shown above.
Difference: Assume a < c. These equivalence classes can be represented by segments
AB ∈ a and AC ∈ c such that A ∗ B ∗ C. Now BC represents the class b such
that a + b = c. The converse is as obvious.
Comparison of Sums The proof is left to the reader.
Proposition 5.8 (Comparison of supplements). Let a segment P Q and two points
A, B in it be given. Assume that P A < P B. Then BQ < AQ.
Proof. The reader has to try himself.
138
5.2
Some elementary triangle congruences
Definition 5.3 (Triangle, Euler’s notation). For a triangle ABC, it is assumed
that the three vertices do not lie on a line. I follow Euler’s conventional notation for
vertices, sides and angles: in triangle ABC, let A, B and C be the vertices, let the
segments a = BC, b = AC, and c = AB be the sides and the angles α := ∠BAC,
β := ∠ABC, and γ := ∠ACB be the angles.
For a segment AB, it is assumed that the two endpoints A and B are different. For
a triangle ABC, it is assumed that the three vertices do not lie on a line.
Proposition 5.9 (Isosceles Triangle Proposition). [Euclid I.5, and Hilbert’s theorem 11] An isosceles triangle has congruent base angles.
Question. Formulate the theorem with specific quantities from a triangle ABC. Provide a drawing.
Answer. If a ∼
= b, then α ∼
= β.
Figure 5.7: An isosceles triangle
Proof. This is an easy application of SAS-congruence. Assume that the sides AC ∼
= BC
are congruent in ABC. We need to show that the base angles α = ∠BAC and
β = ∠ABC are congruent. Define a second triangle A B C by setting
A := B , B := A , C := C
(It does not matter that the second triangle is just ”on top” of the first one.) To apply
SAS congruence, we match corresponding pieces:
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(1) ∠ACB = ∠BCA = ∠A C B because the order of the sides of an angle is arbitrary.
By axiom III.4, last part, an angle is congruent to itself. Hence ∠ACB ∼
= ∠A C B .
(2) AC ∼
= A C .
Question. Explain why this holds.
Answer. AC ∼
= BC because we have assumed the triangle to be isosceles, and
BC = A C by construction. Hence AC ∼
= A C .
(3) Similarly, we show that (3): BC ∼
= B C : Indeed, BC ∼
= AC because we have
assumed the triangle to be isosceles, and congruence is symmetric, and AC = B C by construction. Hence BC ∼
= BC .
Finally, we use axiom III.5. Items (1)(2)(3) imply ∠BAC ∼
= ∠B A C = ∠ABC. But
this is just the claimed congruence of base angles.
The next Proposition is Theorem 12 of Hilbert’s Foundations of Geometry: Here
only the weak form of the SAS-axiom (Hilbert’s Axiom III.5) is assumed.
Question. Why is Hilbert’s Axiom III.5 weaker than the SAS congruence theorem?
Answer. In Hilbert’s Axiom, only congruence of a further pair of angles is postulated.
In the SAS congruence theorem, all pieces of the two triangles are stated to be pairwise
congruent.
Figure 5.8: SAS congruence
Proposition 5.10 (Hilbert’s SAS-axiom implies the full SAS Congruence Theorem). [Theorem 12 in Hilbert] Given are two triangles. We assume that two sides and
the angle between these sides are congruent to the corresponding pieces of the second triangle. Then the two triangles are congruent, which means that all corresponding pieces
are pairwise congruent.
140
Proof. Let ABC and A B C be the two triangles. We assume that the angles at A
and A as well as two pairs of adjacent sides are matched:
c = AB ∼
= ∠C A B = α
= A B = c , b = AC ∼
= A C = b , α = ∠CAB ∼
We need to show that
(a)
(b)
(c)
β = ∠ABC ∼
=∠A B C = β γ = ∠BCA∼
=∠B C A = γ a = BC ∼
=B C = a
Note that by Hilbert’s weaker SAS-axiom, we can concluded only part (a). Part (b)
follows immediately by applying Hilbert’s axiom to triangles ACB and A C B . We
−−→
need still to show part (c). Transferring segment BC onto ray B C , we get a seventh
point D on that ray such that
(1)
BC ∼
= B D
Applying Hilbert’s (weak) SAS axiom to the two triangles ABC and A B D yields
∠BAC ∼
= ∠B A C . By unique= ∠B A D . On the other hand, by assumption ∠BAC ∼
ness of angle transfer (see Hilbert’s axiom III.4), there exists one and only one ray, that
−−→
forms with the ray A B the given ∠BAC, and lies on the same side of the line A B as
−−→
−−→
−−→
point C . Hence rays A D = A C are equal, in other words D lies on the ray A C .
−−→
By construction of A B D , point D lies on the ray B C , too. Since the intersection
point of the two lines A C and B C is unique, one concludes
(2)
C = D
From (1) and (2) we get BC ∼
= B C , as to be shown.
Proposition 5.11 (Extended ASA-Congruence Theorem). Given is a triangle
and a segment congruent to one of its sides. The two angles at the vertices of this side
are transferred to the endpoints of the segment, and reproduced in the same half plane.
Then the newly constructed rays intersect, and one gets a second congruent triangle.
Proof. Let the triangle be ABC and the segment A B ∼
= AB. The angle β = ∠ABC
−−→
is transferred at B along ray B A , and the angle α = ∠BAC is transferred at A along
−−→
ray A B , both are reproduced on the same side of line A B . One gets two new rays rA
and rB such that
−−→ −−→ −→ −−→
−→ −→
) , β = ∠(BA, BC) ∼
)
α = ∠(AB, AC) ∼
= ∠(B A , rB
= ∠(A B , rA
It is claimed that the two new rays rA
and rB
do intersect at some point C . Furthermore, it is claimed that the two triangles ABC and A B C are congruent.
141
Figure 5.9: Extended ASA congruence
On the newly produced ray rA
, we transfer segment AC and get a point C such
that AC ∼
= A C . Now SAS congruence (Theorem 12 of Hilbert, see Proposition 5.10
above) implies
(1)
ABC ∼
= A B C The three pairs of matching pieces used to prove the congruence are stressed in matching
−−→ −−−→
colors. Congruence (1) implies β = ∠ABC ∼
= ∠A B C = ∠(B A , B C ). On the other
Figure 5.10: How to simply get ASA congruence
−−→ hand, ∠ABC ∼
) was assumed, too. Hence by uniqueness of angle transfer,
= ∠(B A , rB
−−−→
. Thus the point C lies on the newly produced ray
we get two equal rays: B C = rB
intersect in point C , and (1) holds, as to be
rB , too. Thus the two rays rA and rB
shown.
Proposition 5.12 (ASA Congruence). [Theorem 13 in Hilbert] Two triangles with
a pair of congruent sides, and pairwise congruent adjacent angles are congruent.
Problem 5.1. State the ASA congruence, with the notation from the figure above. Prove
the theorem.
142
Answer (Independent proof of ASA congruence). Given are the triangles ABC and
A B C , with congruent sides AB ∼
= A B and two pairs of congruent adjacent angles
−−→
at A, A and B, B . The segment AC is transferred onto the ray A C . On this ray, one
gets a point X such that AC ∼
= A X. Now SAS congruence is applied to the triangles
ABC and A B X. The three pairs of matching pieces used to prove the congruence
are stressed in matching colors. By axiom (III.5), one concludes
(5.1)
∠ABC ∼
= ∠A B X
and from the SAS congruence theorem one gets even
(5.2)
ABC ∼
= A B X
We now have obtained both ∠ABC ∼
= ∠A B C as assumed,
= ∠A B X, and ∠ABC ∼
too. Hence by uniqueness of angle transfer, we know the rays to be equal:
−−→ −− →
BX=BC
−−→ −−→
Thus the point X lies on both this ray, and the ray A X = A C , too. These two rays
have a unique intersection point, since they do not lie on the same line. Hence X = C ,
and (5.2) is just the required triangle congruence.
Problem 5.2. In which point does the extended ASA congruence theorem extend the
usual ASA congruence theorem?
Answer. The extended ASA theorem differs from the usual ASA-congruence theorem,
because existence of a second triangle is not assumed—besides the first triangle, only a
second segment is given. It is proved that the two newly produced rays do intersect.
Proposition 5.13 (Preliminary Converse Isosceles Triangle Proposition). If the two
base angles of a triangle are congruent to each other, the triangle is isosceles.
Question. Formulate the theorem with specific quantities from a triangle ABC. Provide a drawing.
Answer. If α ∼
= β and β ∼
= α, then a ∼
= b.
Remark. We can avoid that awkward assumption, ”If α ∼
= β and β ∼
= α”, later in
Proposition 5.32. Hilbert proves the converse isosceles triangle theorem as his Theorem
24, after having the exterior angle theorem at his disposal.
Proof. We use ASA-congruence to prove the Proposition. Assume that the angles α =
∠CAB and β = ∠ABC are congruent in ABC. We need to show that the two sides
a = BC and b = AC are congruent. Define a second triangle A B C by setting
A := B , B := A , C := C
(It does not matter that the second triangle is just ”on top” of the first one.) To apply
ASA congruence, we match corresponding pieces:
143
Figure 5.11: An isosceles triangle, two ways to look at it
(1) AB = B A = A B . Hence AB ∼
= A B , because the order of the endpoints of a
segment is arbitrary, and a segment is congruent to itself.
(2) α ∼
= α .
Question. Explain why this holds.
Answer. α ∼
= β by assumption, and β = ∠ABC = ∠B A C = α by construction.
Hence α ∼
= α .
(3) Similarly, one shows that β ∼
= β : β ∼
= α by assumption, and α = ∠BAC =
∠A B C = β by construction. Hence β ∼
= β .
Via ASA congruence, items (1)(2)(3) imply that ABC ∼
= A B C , and hence espe
cially AC ∼
= A C = BC as to be shown.
5.3
Congruence of angles
Definition 5.4 (Supplementary Angles). Two angles are called supplementary angles, iff they have a common vertex, both have one side on a common ray, and the two
other sides are the opposite rays on a line.
Definition 5.5 (Vertical Angles). Two angles are called vertical angles, iff they have
a common vertex, and their sides are two pairs of opposite rays on two lines.
Proposition 5.14 (Congruence of Supplementary Angles). [Theorem 14 of Hilbert]
If an angle ∠ABC is congruent to another angle ∠A B C , then its supplementary angle
∠CBD is congruent to the supplementary angle ∠C B D of the second angle.
Proof. The three steps of the proof each identify a new pair of congruent triangles.
144
Step 1: One can choose the points A , C and D on the given rays from B such that
AB ∼
= C B , DB ∼
= D B = A B , CB ∼
Because of the assumption ∠ABC ∼
= ∠A B C , SAS congruence (Theorem 12 of Hilbert,
see Proposition 5.10 above) now implies that ABC ∼
= A B C . In the drawing, the
three pairs of matching pieces used to prove the congruence are stressed in matching
colors. Congruence of the two triangles implies
Figure 5.12: Congruence of supplementary angles, the first pair of congruent triangles
(1)
AC ∼
= A C and
∠BAC ∼
= ∠B A C Step 2: By axiom III.3, adding congruent segments yields congruent segments. Hence
the segments AD and A D are congruent. Now SAS congruence (Theorem 12 of Hilbert,
see Proposition 5.10 above) implies that the (greater) triangles, too, are congruent in
the two figures below. In the drawing, the three pairs of matching pieces used to prove
the congruence are stressed in matching colors. The congruence CAD ∼
= C A D
Figure 5.13: Congruence of supplementary angles, the second pair of congruent triangles
implies
(2)
CD ∼
= C D and
∠ADC ∼
= ∠A D C Step 3: At last we consider the two triangles BCD and B C D on the right side.
In the drawing, the three pairs of matching pieces used to prove the congruence are
stressed in matching colors. Again by using SAS congruence (Theorem 12 of Hilbert,
see Proposition 5.10 above), we see the two triangles are congruent. Finally ∠CBD ∼
=
∠C B D , as to be shown.
145
Figure 5.14: Congruence of supplementary angles, the third pair of congruent triangles
Figure 5.15: Supplementary angles yield points on a line
Corollary 5. Adjacent angles congruent to supplementary angles are supplementary,
too.
Proof. Given are supplementary angles ∠ABC and ∠DBC, a further congruent angle
∠ABC ∼
= ∠A B C , and a point D with A and D lying on different sides of line B C .
We shown that the angles ∠A B C and ∠D B C are supplementary if and only if
∠CBD ∼
= ∠C B D . Above, we have already shown one direction: If the angles ∠A B C and ∠D B C are supplementary, then ∠CBD ∼
= ∠C B D .
Now we show the converse. Assume that ∠CBD ∼
= ∠C B D . We have to check
whether point B lies between A and D . Choose any point D on the ray opposite to
−−→
B A . Congruence of supplementary angles (Hilbert’s Theorem 14, see Proposition 5.14
above) implies ∠CBD ∼
= ∠C B D . On the other hand, ∠CBD ∼
= ∠C B D is assumed.
−−→
Angle ∠CBD is transferred uniquely along ray B C into the half plane opposite to A .
−−−→ −−→
Indeed, by axiom III.4, angle transfer produces a unique new ray. Hence B D = B D .
and the four points A , B , D , D lie on one line, with point B between A and D .
Hence angles ∠A B C and ∠D B C are supplementary, too.
Proposition 5.15 (Congruence of Vertical Angles). [Euclid I.15] Vertical angles
are congruent.
Proof. This is an easy consequence of Hilbert’s Theorem 14 about supplementary angles
(see Proposition 5.14 above). Take any two vertical angles ∠ABC and ∠A BC . We
assume that vertex B lies between points A and A on one line, as well as between
the two points C and C on a second line. As shown in the figures, angle ∠ABC has
angle (i) ∠ABC as supplementary angle. Secondly, angle ∠ABC has (ii) ∠A BC as
supplementary angle, too.
An angle is congruent to itself, as stated in Axiom III.4. Hence especially ∠ABC ∼
=
∠ABC . By Theorem 14 of Hilbert (see Proposition 5.14 above), angles supplementary
146
Figure 5.16: Two pairs of supplementary angles yield vertical angles
to congruent angles are congruent, too. Hence we conclude congruence of the two vertical
angles: ∠ABC ∼
= ∠A BC , as to be shown.
Proposition 5.16 (The main case of angle-addition). Given is an angle ∠ABC
−−→
−−→
and a ray BG in its interior, as well as a second angle ∠A B C and a ray B G in its
interior. Furthermore, assume that ∠CBG ∼
= ∠A B G . Then
= ∠C B G and ∠ABG ∼
the two angle sums are congruent, too: ∠ABC ∼
= ∠A B C .
Figure 5.17: Angle addition
Proof. By the Crossbar Theorem, a segment going from one side of an angle to the
other, and a ray in the interior of that angle always intersect. Hence there exists a point
−−→
−−→
H such that BG = BH and A H C. For simplicity, we choose G = H from the
beginning. Too, we may assume that A , C and G are chosen such that BA ∼
= B A ,
∼
∼
BC = B C and BG = B G . The proof uses three pairs of congruent triangles, in step
(1)(2)(3), respectively.
Step (1): The SAS-congruence axiom implies BGC ∼
= B G C because of
∠GBC ∼
= ∠G B C , GB ∼
= G B and BC ∼
= B C which hold by assumption and the
147
remarks above. In the drawing, the three pairs of matching pieces used to prove the
congruence are stressed in matching colors. From the congruence of these two triangles,
Figure 5.18: the first pair of congruent triangles
we conclude that
(1)
(1b)
∠BCG ∼
= ∠B C G
∠BGC ∼
= ∠B G C Step (2): As a second step, the SAS-congruence axiom implies BGA ∼
= B G A
because of ∠GBA ∼
= ∠G B A , GB ∼
= G B and BA ∼
= B A which hold by assumption
and the remarks above. Again, in the drawing, the three pairs of matching pieces used
to prove the congruence are stressed in matching colors. By construction, the two angles
Figure 5.19: the second pair of congruent triangles
∠BGA and ∠BGC are supplementary angles. The congruence ∠BGA ∼
= ∠B G A follows from step (2), and ∠BGC ∼
= ∠B G C as stated by (1b). As derived in our Corollary
to Hilbert’s Theorem 14 (see Proposition 5.14 above), adjacent angles congruent to supplementary angles are supplementary, too. Hence the two angles ∠B G A and ∠B G C 148
are supplementary. Hence the three points A , G and C lie on a straight line. By
segment addition (Hilbert’s axiom III.3), AG ∼
= A G and GC ∼
= G C imply
(2)
AC ∼
= A C Step (3): To set up a third pair of congruent triangles, we use (1) (2) and the assumption
(3)
BC ∼
= BC The SAS axiom shows that ABC ∼
= A B C because of (1)(2)(3). I have stressed
these three pairs of matching pieces in matching colors. Finally, ∠ABC ∼
= ∠A B C Figure 5.20: the third pair of congruent triangles
follows from ABC ∼
= A B C , as to be shown.
Proposition 5.17 (The main case of angle subtraction). Given is an angle ∠ABC
−−→
−−→
and a ray BG in its interior, as well as a second angle ∠A B C and a ray B G in its
interior. If ∠CBG ∼
= ∠C B G and ∠ABC ∼
= ∠A B C then ∠ABG ∼
= ∠A B G .
Proposition 5.18 (Angle-Addition and Subtraction). [Theorem 15 in Hilbert]
Given are three vertices h, k, l with the common vertex O lying in a plane a, and three
vertices h , k , l with the common vertex O lying in a plane a . We assume either the
case of angle subtraction or angle addition.
angle subtraction: in this case the rays h and k lie in the same half-plane of l, and
the rays h and k lie in the same half-plane of l ;
angle addition: in this case the rays h and k lie in the different half-planes of l, and
the rays h and k lie in different half-planes of l .
−−→
Given is an angle ∠ABC and a ray BG in its interior, as well as a second angle ∠A B C −−→
and a ray B G in its interior. If
∠(h, l) ∼
= ∠(h , l )
and
149
∠(l, k) ∼
= ∠(l , k )
then either both h and k are opposite rays, and h and k are opposite rays, too; or
∠(h, k) ∼
= ∠(h , k )
Proof. We explain only the case of angle addition. In the special case that h and k
are opposite rays, the angles ∠(h, l) and ∠(l, k) are adjacent supplements. Hence by
Corollary 5, the congruent angles ∠(h , l ) and ∠(l , k ) are supplementary, too. In other
words we get opposite rays h and k . We have to distinguish the two remaining main
cases:
(i) the two rays k and l lie in one half-plane of ray h;
(ii) the two rays k and l lie in different half-planes of ray h.
In the first case (i), the two rays k and l lie in one half-plane of ray h . The result
follows by angle addition as in Proposition 5.16.
Here is a detailed argument: We transfer angle ∠(h, k) onto the ray h in the half
plane determined by l and get a ray k+
such that ∠(h, k) ∼
). The Proposi= ∠(h , k+
∼
tion 5.17 about the main case of angle subtraction implies ∠(l.k) = ∠(l , k+
). Because
∼
of the assumption ∠(l.k) = ∠(l , k ), uniqueness of angle transfer implies k+ = k Hence
the two rays k and l lie in one half-plane of ray h .
from l . Now
In the second case (ii), we use the opposition rays l− of l and l−
Proposition 5.14 about the congruence of supplementary angles implies
∠(h, l− ) ∼
)
= ∠(h , l−
and
∠(l− , k) ∼
, k)
= ∠(l−
Since we are back to case (i), we can conclude the claim ∠(h, k) ∼
= ∠(h , k ).
−−→
Definition 5.6 (The sum of two angles). Let an angle ∠ABC and a ray BG in its
interior. Angle ∠ABC is called the sum of the angles ∠ABG and ∠GBC. One writes
∠ABC = ∠ABG + ∠GBC.
Figure 5.21: Three cases for adding angles: they (i) are supplementary
(iii) or cannot be added
150
(ii) can be added
Lemma 5.2. Given are two angles ∠ABG and ∠GBC with same vertex B lying on
−−→
different sides of a common ray BG. Exactly one of three possibilities occur about the
angles ∠ABG and ∠GBC:
(i) They are supplementary. The three points A, B and C lie on a line.
(ii) They can be added. Their sum is ∠ABG + ∠GBC = ∠ABC. Points A and G
lie on the same side of line BC. Points C and G lie on the same side of line AB.
(iii) They cannot be added. Points A and G lie on different sides of line BC.
Points C and G lie on different sides of line AB. (The sum would be an overobtuse angle.)
Figure 5.22: If A and G lie on different sides of BC, then C and G lie on different sides of
AB.
Proof. Suppose that neither case (i) nor (ii) occurs. Under that assumption, either
(a) points A and G lie on different sides of line BC—or
(b) points C and G lie on different sides of line AB.
Suppose case (a) occurs. Segment AG intersects line BC, say in point H. We use
Pasch’s axiom for triangle CHG and line AB. This line does not intersect side GH,
but intersects side CH. Indeed, point B lies between C and H, since C and A, and
hence C and H are on different sides of line BG.
Now Pasch’s axiom—for triangle CHG and line AB—implies that this line intersects side CG, say at point P asch. Thus points C and G lie on different sides of line
AB. We have shown case (iii) to occur.
Suppose case (b) occurs. The same argument—with A and C exchanged— shows
that case (iii) occurs once more.
151
Remark. Because of these theorems, angle addition and subtraction are defined for
congruence classes of angles.
152
5.4
SSS congruence
The ”Theorem about the symmetric kite” is needed as a preparation to get SSS congruence.
Proposition 5.19 (The Symmetric Kite). [Theorem 17 of Hilbert] Let Z1 and Z2 be
two points on different sides of line XY , and assume that XZ1 ∼
= XZ2 and Y Z1 ∼
= Y Z2 .
Under these assumptions
(i) the angles ∠XZ1 Y ∼
= ∠XZ2 Y are congruent;
(ii) the angles ∠XY Z1 ∼
= ∠XY Z2 are congruent.
Proof. The congruence of the base angles of isosceles XZ1 Z2 yields ∠XZ1 Z2 ∼
= ∠XZ2 Z1 .
Similarly, one gets ∠Y Z1 Z2 ∼
= ∠Y Z2 Z1 . Now angle addition (or substraction) implies
(*)
∠XZ1 Y ∼
= ∠XZ2 Y
−−−→
Angle addition is needed in case ray Z1 Z2 lies inside ∠XZ1 Y , angle subtraction in case
−−−→
ray Z1 Z2 lies outside ∠XZ1 Y .
In the special case that either point X or Y lies on the line Z1 Z2 , one gets the same
conclusion even easier. Here are drawings for the three cases. One now applies SAS
Figure 5.23: The symmetric kite
congruence to the triangles XZ1 Y and XZ2 Y . Indeed, the angles at Z1 and Z2
and the adjacent sides are pairwise congruent. Hence the assertion ∠XY Z1 ∼
= ∠XY Z2
follows.
Before getting the general SSS-congruence, I consider one further special case.
Lemma 5.3 (Lemma for SSS Congruence). Assume that the two triangles ABC
and AB C have a common side AC, and all three corresponding sides are congruent,
and the two vertices B and B lie on the same side of line AC. Then the two triangles
are identical.
153
−→
Proof. We transfer the angle ∠BAC onto the ray AC, on the side of line AC opposite
to B and B . On the newly produced ray, we transfer segment AB, starting at vertex
A. Thus we get point B , and segment AB ∼
= AB. From SAS-congruence (Theorem
12 of Hilbert, see Proposition 5.10 above), one concludes that BC ∼
= B C. As stressed
in the first drawing, one has constructed a symmetric kite with the four points
X := A, Y := C, Z1 := B, Z2 := B Hence by Theorem 17 (see Proposition 5.19 above), ∠B AC ∼
= ∠BAC
Figure 5.24: Which kite is symmetric?
But wait! Another choice of four points to get a kite is
X := A, Y := C, Z1 := B , Z2 := B —replacing B by B . Because of the assumptions, and the construction of the first kite,
AB ∼
= AB ∼
= AB and B C ∼
= BC ∼
= B C. We see that the second kite, stressed in the
second drawing, satisfies the assumptions of Hilbert’s Theorem 17 (see Proposition 5.19
above), too. Now we conclude from Hilbert’s Theorem 17 (see Proposition 5.19 above)
−→
that ∠B AC ∼
= ∠B AC. Finally, the uniqueness of angle transfer implies that AB =
−−→
AB . Since AB ∼
= AB , and segment transfer was shown to produce a unique point, we
have confirmed that B = B . Thus the two triangles ABC and AB C are identical,
as to be shown.
Remark. We do not need to assume that angle congruence is an equivalence relation.
But therefore we have to consider two kites, by choosing two of the three triangles, as
shown in two drawings. Then we can use that transferring an angle gives a unique ray.
Proposition 5.20 (The diagonals of the rhombus bisect each other perpendicularly). Given are four different points A, B, C, D such that the segments AB ∼
=
∼
∼
BC = CD = DA are congruent. Then the segments AC and BD bisect each other
perpendicularly at their common midpoint.
154
Figure 5.25: The diagonals of the rhombus bisect each other.
Answer. Points A and C lie on different sides of line BD. Otherwise the lemma for
SSS-congruence would imply A = C, contrary to the assumption that four different
points A, B, C, D are given. Similarly, we see that points B and D lie on different sides
of the other diagonal AC. Hence the two segments AC and BD intersect. Let point M
be their intersection point.
We draw the horizontal diagonal AC. The upper- and lower triangles have two pairs
of congruent bases angles (see second figure). Angle addition implies ∠ABD ∼
= ∠CBD.
By SAS congruence, the left- and right triangles ABD ∼
= CBD are congruent (see
third figure).
Furthermore, the latter two triangles are both isosceles. Hence ABD ∼
= DCB
holds, too. Again by SAS congruence, we get four mutually congruent base angles with
one side on the line BD (first figure in the second row). Similarly, we get four mutually
congruent base angles with one side on the line AC.
By ASA congruence, we get ABM ∼
= CDM . Hence the diagonals AC and BD
bisect each other (see last figure in the second row).
Another ASA congruence yields AM B ∼
= CM B. Hence the angles ∠AM B ∼
=
∠CM B are congruent supplements, and hence they are right angles. Hence the diagonals
AC and BD are perpendicular to each other.
From Hilbert’s Theorem 17 (see Proposition 5.19 above), and the Lemma 5.3 above,
155
we see that SSS-congruence holds for any two triangles with a common side. Now we
can easily get the general case of
Proposition 5.21 (SSS Congruence). [Theorem 18 in Hilbert’s Foundations] Two
triangles with three pairs of congruent sides are congruent.
Figure 5.26: Which two triangles are congruent?
Proof. Assume the two triangles ABC and A B C have corresponding sides which
−−→
are congruent. We transfer the angle ∠BAC onto the ray A C , at vertex A , to the
same side of A C as B . On the newly produced ray, we transfer segment AB, starting
at vertex A . Thus we get point B0 , such that A B0 ∼
= AB. Because of SAS-congruence
(Theorem 12 of Hilbert, see Proposition 5.10 above), we get
ABC ∼
= A B0 C (*)
Hence especially, AB ∼
= A B and BC ∼
= B C . We can now apply
= A B 0 ∼
= B0 C ∼
the Lemma‘5.3 to the two triangles A B C and A B0 C . Hence B = B0 , and the
assertion follows from (*).
Corollary 6 (Transfer of a triangle). Any given triangle ABC can be transferred
into both the left as well as right half-plane of any given ray r, to obtain a congruent
triangle with one vertex at the vertex of this ray, one side lying on the given ray, and
the third vertex in the prescribed half-plane.
5.5
The equivalence relation of angle congruence
Proposition 5.22 (Theorem 19 in Hilbert’s Foundations). If two angles ∠(h , k )
and ∠(h , k ) are congruent to a third angle ∠(h, k), then the two angles ∠(h , k ) and
∠(h , k ) are congruent, too.
Proof for a bottle of wine from Hilbert to A. Rosenthal. 24 Let the vertices of the angles
be O, O , O . Choose points A, A , A on one side of the three angles, respectively,
such that O A ∼
= OA and O A ∼
= OA. Similarly choose points B, B , B on the
24
Hilbert gives credit for this proof to A. Rosenthal (Math. Ann. Band 71)
156
three remaining sides, respectively, such that O B ∼
= OB and O B ∼
= OB. Now the
assumption of SAS congruence (Theorem 12 of Hilbert, see Proposition 5.10 above) are
met for both A O B and AOB, as well as A O B and AOB. Hence
A B ∼
= AB , A B ∼
= AB
We use axiom III.2:
"two segments congruent to a third one are congruent to each other".
Hence A B O and A B O have three pairs of congruent sides. By the SSScongruence Theorem 18, we conclude that ∠(h , k ) ∼
= ∠(h , k ), as to be shown.
Proposition 5.23 (Congruence is an equivalence relation). Congruence is an
equivalence relation on the class of angles.
Question. Which three properties do we need to check for a congruence relation?
Answer. For an arbitrary relation to be a congruence relation, we need to check
(a) reflexivity: Each angle is congruent to itself, written α ∼
= α.
(b) symmetry: If α ∼
= β, then β ∼
= α.
(c) transitivity: If α ∼
= β, and β ∼
= γ, then α ∼
= γ.
Question. How can we claim reflexivity?
Answer. Reflexivity is given by the last part of axiom III.4
Proof of symmetry, for a bottle of wine from A. Rosenthal to Hilbert. 25 By Theorem 19
(see Proposition 5.22 above), ∠(h , k ) ∼
= ∠(h, k) and ∠(h , k ) ∼
= ∠(h, k) imply ∠(h , k ) ∼
=
∼
β
and
α
β
imply
∠(h , k ). Hence, with just other notation, we see that β ∼
=
=
β∼
= α.
Proof of transitivity. Assume α ∼
= β and β ∼
= γ. Because of symmetry γ ∼
= β. Now we
use Theorem 19 (see Proposition 5.22 above). Hence, just with other notation, α ∼
=β
∼
∼
and γ = β imply α = γ.
Definition 5.7 (Angle comparison). Given are two angles ∠BAC and ∠B A C . We
−−→
say that ∠BAC is less than ∠B A C , iff there exists a ray A G in the interior of ∠B A C such that ∠BAC ∼
= ∠B A G. In this case, we also say that ∠B A C is greater than
∠BAC. We write ∠B A C > ∠BAC and ∠BAC < ∠B A C , equivalently.
Proposition 5.24 (Angle comparison holds for congruence classes). Assume
that α ∼
= α and β ∼
= β . α < β if and only if α < β .
Proof. The reader should do it on her own.
25
Hilbert gave credit and got back a good bottle
157
Proposition 5.25 (Transitivity of angle comparison). If α < β and β < γ, then
α < γ.
Proof. After having done some transfer of angles, I can assume that all three angles
−→
α, β, γ have the common side AB, and lie on the same side of AB. Choose any point E
on the second side of γ = ∠BAE.
Because β < γ, the second side of β is in the interior of the largest angle γ. Hence,
by the Crossbar Theorem the segment BC intersects that second side of β, say at point
D, and β = ∠BAD as well as B ∗ D ∗ E.
Because α < β, the second side of α is in the interior of β . Hence, by the
Crossbar Theorem the segment BD intersects that second side of α, say at point E, and
α = ∠BAE as well as B ∗ C ∗ D. Any four points on a line can be ordered in a way
Figure 5.27: Transitivity of comparison of angles
that all four alphabetic order relations hold. (see Theorem 5 in Hilbert, which I call
the ”four-point Theorem”). Now the four points B, C, D, E satisfy the order relations
B ∗ C ∗ D and B ∗ D ∗ E. Therefore they are already put in alphabetic order. Hence
B ∗ C ∗ E . This shows by definition that α = ∠BAC < ∠BAE = γ.
Proposition 5.26 (All angles are comparable). For any two angles α and β, one
and only one of the three cases (i)(ii)(iii) occurs:
Either (i) α < β
or (ii) α ∼
or (iii) β < α.
=β
Proof. Not more than one of the cases (i)(ii)(iii) can occur at once. Case (ii) excludes
either (i) or (iii) by definition of angle comparison. But case (i) and (iii) cannot hold
both at the same time, neither: By transitivity, α < β and β < α together would imply
α < α. This is impossible, because an angle is congruent to itself: α ∼
= α by axiom III.4.
Now we show that actually one of the cases (i)(ii)(iii) does occur. Transfer angle α
onto one side of β. Thus we get the two angles α
= ∠BAC ∼
= α and β = ∠BAD which
−→
−→
have the common side AB, and lie on the same side of AB. The second side AC of
angle α
can
either (i) lie in the interior of angle β.
158
or (ii) be identical to the second side of β.
or (iii) lie in the exterior of angle β.
These cases correspond to the three cases as claimed. An explanation may be needed in
Figure 5.28: All angles are comparable
−→
case (iii): The ray AC lies in the exterior of angle β = ∠BAD. What does that mean?
Answer. Either points B and C lie on different sides of line AD, or points C and D lie
on different sides of line AB.
The points C and D lie on the same side of AB by the arrangement of the angles. Hence
B and C lie on different sides of AD. By the Crossbar theorem, segment BC intersects
−−→
ray AD. In term to comparing angles, we get β = ∠BAD < ∠BAC = α
∼
= α.
Proposition 5.27 (Comparison of supplements). If α < β, then their supplements
S(α) and S(β) satisfy S(α) > S(β).
Figure 5.29: Comparison of supplements
159
Proof. Transfer angle α onto one side of β, in the same half plane. Thus we get the two
−→
angles α
∼
= α and β = ∠BAD which have the common side AB, and lie on the same
side of AB.
Because of α < β, the second side of angle α
lies in the interior of angle β. By the
Crossbar theorem it intersects the segment BD, say at point Q. We have arranged that
−→
α ∼
= ∠BAQ, and β = ∠BAD. Let F be any point on the ray opposite to AB.
= α
Apply Pasch’s axiom (Hilbert’s axiom of order II.4) to F BQ and line AD. That line
does not intersect side BQ because of B ∗ Q ∗ D, but does intersect side F B because of
F ∗ A ∗ B. Hence line AD intersects the third side F Q, say at point S. Indeed F ∗ S ∗ Q,
−−→
and S lies on the ray AD because only that ray, and not its opposite ray lies in the
interior of ∠F AQ.
The supplementary angle of β is S(β) = ∠F AD = ∠F AS. The supplementary angle
of α
is S(
α) = ∠F AQ. From F ∗ S ∗ Q we get
S(β) = ∠F AS = ∠F AD < ∠F AQ = S(
α)
The final step uses Hilbert’s Theorem 14 (see Proposition 5.14 above): the supplements
of congruent angles are congruent, too. Hence α ∼
implies S(α) ∼
α). From
= α
= S(
∼
S(β) < S(
α) and S(α) = S(
α), we conclude S(β) < S(α) as to be shown.
Definition 5.8 (acute, right and obtuse angles). A right angle is an angle congruent
to its supplementary angle. An acute angle is an angle less than a right angle. An obtuse
angle is an angle greater than a right angle.
Remark (Remark about supplements of acute and obtuse angles). If an angle α is acute,
its supplement S(α) is obtuse. Furthermore, the angle is less than its supplement.
If an angle β is obtuse, its supplement S(β) is acute. Furthermore, the angle is less than
its supplement.
Reason. Assume α < R, where R denotes a right angle. From the comparison of
supplements done in Proposition 5.27, we conclude S(α) > S(R). But by the definition
of a right angle R ∼
= S(R). Hence S(α) > S(R) ∼
= R, and because comparison is
for congruence classes, we get S(α) > R. Hence α < R < S(α), and by transitivity
(Problem 10.2), we conclude α < S(α).
Similarly, we explain that β > R implies S(β) < R and S(β) < β.
Proposition 5.28 (All right angles are congruent).
Question. How is a right angle defined?
Answer. A right angle is an angle congruent to its supplementary angle.
Proof in the conventional style. Let α be a right angle, and β = S(α) ∼
= α be its
congruent supplement. Similarly, we consider a second pair of right angles α and
β = S(α ) ∼
= α being its congruent supplement.
160
The question is whether both right angles α and α are congruent to each other. By
Proposition 5.26 all angles are comparable. Hence the two angles α and α satisfy just
one of the following relations.
Either (i) α < α
or (ii) α ∼
or (iii) α < α.
= α
We need to rule out cases (i) and (iii) by deriving a contradiction. I only need to explain
case (i), because (iii) is similar. Now assume α < α towards a contradiction. By the
comparison of supplements from Proposition 5.27, we get S(α) > S(α ). Because a right
angle is congruent to its supplement we get α ∼
= S(α) > S(α ) ∼
= α . As explained in
Proposition 5.24, angle comparison holds for congruence classes. Hence we get α > α .
Thus, from the assumption α < α , we have derive α > α . Transitivity would imply
α < α which is impossible. (Axiom III.4 states α ∼
= α.)
Thus case (i) leads to a contradiction. Similarly, case (iii) leads to a contradiction.
The only remaining possibility is case (ii): Any two given right angles α and α are
congruent.
Proposition 5.29 (About sums of angles). The sum of angles is defined on equivalence classes of congruent angles. It satisfies the following properties:
Commutativity If α + β exists, then β + α exists and β + α ∼
= α + β.
Associativity (α + β) + γ ∼
= α + (β + γ) and existence of one side implies existence
of the other one.
Difference α < γ if and only if there exists β such that α + β ∼
= γ.
Comparison of Sums 1 If α ∼
= β and γ < δ and β + δ exists, then α + γ exists and
α + γ < β + γ.
Comparison of Sums 2 If α < β and γ < δ and β + δ exists, then α + γ exists and
α + γ < β + γ.
Proof. By Hilbert’s Theorem 15 and Proposition 5.23, the sum of angles is defined on
equivalence classes of congruent angles. Now we check the items stated:
Commutativity: Let α ∼
= ∠ABG and β ∼
= ∠GBC where points G and C lie both on
the same side of line AB. With that setup, α + β ∼
= ∠ABC, which is assumed to
exist. Since the order of the two rays of an angle is defined to be chosen arbitrarily,
∠ABC = ∠CBA ∼
= ∠CBG + ∠GBA ∼
=β+α
hence the latter angle exists and β + α ∼
= α + β.
Associativity: The proof is left to the reader.
161
Difference: Let α ∼
= ∠ABG and γ ∼
= ∠ABC where points G and C lie both on the
−−→
same side of line AB. With that setup, α < γ iff the ray BG lies inside the angle
∠ABC iff α + β ∼
= γ with β ∼
= ∠GBC.
Comparison of Sums 1: Assuming that α ∼
= β and γ < δ and β + δ exists, we know
there exists ε such that γ + ε ∼
= δ. Hence
(α + γ) + ε ∼
= α + (γ + ε) ∼
=β+δ
and α + γ < β + γ where the former is shown to exist.
Comparison of Sums 2: Assuming α < β and γ < δ, we know there exist angles η
and ε such that α + η ∼
= β and γ + ε ∼
= δ. Hence
(α + γ) + η + ε ∼
= (α + η) + (γ + ε) ∼
=β+δ
and α + γ < β + γ where the former is shown to exist.
Proposition 5.30 (The Hypothenuse Leg Theorem). Two right triangles for which
the two hypothenuse, and one pair of legs are congruent, are congruent.
Question. Of which one of SAS, SSA, SAA, ASA, SSS congruence contains the hypothenuseleg theorem as a special case? Is there a corresponding unrestricted congruence theorem?
Answer. The hypothenuse-leg theorem is a special case of SSA congruence. There is no
unrestricted SSA congruence theorem.
First proof of the hypothenuse-leg theorem 5.30 . Given are two right triangles ABC
and
A B C . As usual, we put the right angles at vertices C and C . We assume congruence
of the hypothenuses AB ∼
= A C . We can build a kite
= A B and of one pair of legs AC ∼
out of two copies of triangle ABC and two copies of triangle A B C .
−→
To this end, one transfers angle ∠C A B onto the ray AC into the opposite half
plane and gets ∠C A B ∼
= ∠CAB . Point B can be chosen on the newly produced ray
such that A B ∼
= AB . By SAS-congruence it is easy to confirm A B C ∼
= AB C
and especially
(5.3)
BC ∼
= B C
Too, the two supplementary right angles at vertex C imply that points B, C and B lie
on a line. (Why?)
−→
Next we transfer segment CA onto the ray opposite to CA and get point D such
that CA ∼
= CD. We produce the two triangles in the right half of the figure. Hence SAS
congruences using the right angle imply ABC ∼
= DBC and AB C ∼
= DB C.
162
Since AB ∼
= A B ∼
= AB ∼
= AB and DB ∼
= DB , one has constructed a
= AB ∼
symmetric kite with the four points
X := A, Y := D, Z1 := B, Z2 := B as shown in the drawing. We can apply Hilbert’s kite-theorem 5.19 and conclude
Figure 5.30: Four right triangles yield a kite. When is it even a rhombus?
∠DAB = ∠XY Z1 ∼
= ∠XY Z2 = ∠DAB The congruence of the same angles is ∠CAB = ∠B AC. Hence SAS congruence implies
now CAB ∼
= B AC and
BC ∼
= B C
Together with formula (5.3) above we conclude B C ∼
= BC. Now a final SAS congruence
using the right angles implies implies congruence ABC ∼
= A B C of the originally
given triangles, as to be shown.
5.6
Constructions with Hilbert tools
Now we show that a right angle actually exists, and do some further basic constructions
with the Hilbert tools. The only means of construction are those granted by Hilbert’s
axioms.
Definition 5.9 (Hilbert tools). By incidence axiom I.1 and I.2, we can draw a unique
line between any two given points. By axiom III.1 and Proposition 5.2 from the very
163
beginning, we transfer a given segment uniquely to a given ray. Finally by axiom
III.4, we transfer a given angle uniquely on a given ray into the specified half plane.
Constructions done using only these means are called constructions by Hilbert tools.
Problem 5.3 (Drop a Perpendicular). Given is a line OA and a point B not on
this line. We have to drop the perpendicular from point B onto line OA.
−−→
Construction 5.1. Draw ray OB. Transfer angle ∠AOB, into the half plane opposite
−→
to B, with ray OA as one side. On the newly produced ray, transfer segment OB to
produce a new segment OC ∼
= OB. The line BC is the perpendicular, dropped from
point B onto line OA.
Figure 5.31: Drop the perpendicular, the two cases
Proof of validity. The line OA and the segment BC intersect, because B and C lie on
different sides of OA. I call the intersection point M . It can happen that O = M . In
that special case A = M and
(**)
∠AM B ∼
= ∠AM C
But these is a pair of congruent supplementary angles, because M lies between B and
C. By definition, an angle congruent to its supplementary angle is a right angle. Hence
∠AM B is a right angle.
In the generic situation O = M , we distinguish two cases
(i) On the given line OA, points M and A lie on the same side of O.
(ii) Point O lies between M and A.
164
In both cases we show by the SAS-congruence theorem that the two triangles OM B
and OM C are congruent. Indeed, the angles at vertex O are congruent, both in case
(i) and (ii):
−→ −−→
In case (i) , the rays OA = OM are equal, and ∠M OB = ∠AOB ∼
= ∠AOC = ∠M OC
follows from the construction.
−→
−−→
In case (ii) , the rays OA and OM are opposite. Thus ∠M OB and ∠AOB, as well as
∠M OC and ∠AOC are supplementary angles. Again ∠AOB ∼
= ∠AOC because
of the angle transfer done in the construction. Now Theorem 14 of Hilbert (see
Proposition 5.14 above) tells that supplements of congruent angles are congruent.
Hence we get ∠M OB ∼
= ∠M OC once again.
The adjacent sides OB ∼
= OC are congruent by construction, too. Finally the common
sides OM is congruent to itself. The drawing stresses the pieces matched to prove the
congruence. Because the two triangles are congruent, the corresponding angles at vertex
Figure 5.32: Congruences needed in the two cases
M are congruent, too. Hence
(*)
∠OM B ∼
= ∠OM C
which is a pair of congruent supplementary angles. A right angle is, by definition, an
angle congruent to its supplementary angle . Hence either ∠OM B is a right angle
because of formula (*), or, in the special case O = M , angle ∠AM B is a right angle
because of (**).
Proposition 5.31 (Supplements of acute and obtuse angles). Let R denote a
right angle. For any angle γ and its supplement S(γ) exactly one of the following three
cases occurs: Either (1) or (2) or (3).
165
(1) γ < R, S(γ) > R and γ < S(γ)
(2) γ ∼
= R, S(γ) ∼
= R and γ ∼
= S(γ)
(3) γ > R, S(γ) < R and γ > S(γ).
Proof. All angles are comparable (see Proposition 5.26). For angle γ and the right angle
R, exactly one of the three cases holds:
Either (i) γ < R
or (ii) γ ∼
or (iii) γ > R.
=R
From SAS congruence (Theorem 12 of Hilbert, see Proposition 5.10 above), we know that
α∼
= β implies S(α) ∼
= S(β). From Proposition 5.27 about comparison of supplements,
we know that α < β implies S(α) > S(β). Hence with the help of Propositions 5.23
and 5.24 we get:
in case (i), one concludes S(γ) > S(R) ∼
= R > γ, and hence (a) γ < S(γ),
in case (ii), one concludes S(γ) ∼
= S(R) ∼
=R∼
= γ, and hence (b) γ ∼
= S(γ),
in case (iii), one concludes S(γ) < S(R) ∼
= R < γ, and hence (c) γ > S(γ).
By Proposition 5.26 above, all angles are comparable. Hence the two angles γ and its
supplement S(γ) satisfy either (a) or (b) or (c). This observation allows one to get the
converse of the conclusions shown above. For example, (a) excludes both (ii) and (iii),
and hence implies (i). Thus we get
(a) implies (i) , (b) implies (ii) , (c) implies (iii);
and finally
(a) if and only if (i) , (b) if and only if (ii) , (c) if and only if (iii).
This leads to the mutually exclusive cases (1) (2) (3), as originally stated.
Corollary 7 (All right angles are congruent).
Proof. The proposition 5.31 about supplements of acute and obtuse angles yields an
easy proof that all right angles are congruent:
Let R be the right angle as has been constructed in Construction 5.1. Now suppose
γ = R is another right angle. By definition of a right angle, this means that γ ∼
= S(γ).
∼
Hence case (b) above occurs, which implies (ii): γ = R. You see that existence of a
right angle make proving its uniqueness a bit easier.
Proposition 5.32 (Converse Isosceles Triangle Proposition). [Euclid I.6, Theorem 24 of Hilbert] A triangle with two congruent angles is isosceles.
Question. Formulate the theorem with specific quantities from a triangle ABC.
Answer. If α ∼
= β, then a ∼
= b.
166
Proof. Given is a triangle ABC with α ∼
= β. By Hilbert’s Theorem 19 (see Proposition 5.22 above) and Proposition 5.23, congruence of angles is an equivalence relation.
Hence α ∼
= β implies β ∼
= α. In the preliminary version given as Proposition 5.13, we
∼
have shown that α = β and β ∼
= α together imply a ∼
= b. Hence the ABC is isosceles,
as to be shown.
Proposition 5.33 (Existence of an Isosceles Triangle). For any given segment
AB, on a given side of line AB, there exists an isosceles triangle.
Of course, this triangle is not unique.
Construction 5.2 (Construction of an isosceles triangle). Given is a segment
AB. Choose any point P not on line AB, in the half plane specified. Compare the two
angles ∠BAP and ∠ABP . In the drawing, ∠BAP is the smaller angle. Transfer the
−→
smaller angle, with ray BA as one side, and the newly produced ray rB in the same half
plane as P . Ray rB and segment AP do intersect. The intersection point C lies in the
half plane as required, and ABC is isosceles.
Figure 5.33: Construction of an isosceles triangle
Proof of validity. By Proposition 5.26, all angles are comparable. Comparison of the
two angles ∠BAP and ∠ABP leads to one of the three possibilities:
Either (i) The two angles are congruent
or (ii) α = ∠BAP < ∠ABP = β
or (iii) α = ∠BAP > ∠ABP = β.
167
It is enough to consider cases (i) and (ii). In case (i) let C = P . In case (ii) we have
−→
transferred the smaller angle ∠BAP , onto ray BA, and the newly produced ray rB in
the same half plane as P . By the Crossbar Theorem, a segment going from one side
of an angle to the other, and a ray in the interior of an angle always intersect. By the
Crossbar Theorem, ray rB and segment AP intersect. The intersection point C lies
between A and P , hence both points C and P lie in the same half plane of line AB.
Hence point C lies in the half plane as required.
By construction, the base angles of triangle ABC are congruent. Hence the triangle
is isosceles by Euclid I.6 (Converse Isosceles Triangles).
Remark. One may suggest to make it part of the construction how to compare the two
angles ∠BAP and ∠ABP . One possibility is to transfer both angles to the other vertex,
B or A, respectively. Only the newly produced ray from the smaller angle does intersect
the opposite side of ABP .
Problem 5.4 (Erect a Perpendicular). Given is a line l and a point R on this line.
We have to erect the perpendicular at point R onto line l.
Construction 5.3. Choose any point A = R on line l. Transfer segment AR onto
−→
the ray opposite to RA to get a segment RB ∼
= AR. As explained in Construction 5.2,
construct any isosceles triangle ABC. The line CR is the perpendicular to the given
line l through point R.
Figure 5.34: Erect a perpendicular
Reason for validity. The two triangles RAC and RBC
gruence. Indeed, ∠RAC ∼
= ∠RBC by the construction of
thermore, we have a pair of congruent adjacent sides: AR
above, and AC ∼
= BC because ABC is isosceles.
168
are congruent by SAS conthe isosceles triangle. Fur∼
= BR by the construction
Now the triangle congruence RAC ∼
= RBC implies that ∠ARC ∼
= ∠BRC. But
these two angles are supplementary angles. Because congruent supplementary angles
are right angles, we have confirmed that ∠ARC is a right angle, as to be shown.
Definition 5.10 (The perpendicular bisector). The line through the midpoint of a
segment and perpendicular to the segment is called the perpendicular bisector.
Problem 5.5. Given is any segment AB. Construct the perpendicular bisector.
Construction 5.4. Construct two different isosceles triangles over segment AB. The
line connecting the third vertices of the two isosceles triangles is the perpendicular bisector.
Figure 5.35: The perpendicular bisector via the kite
Proof of validity. Let ABC be the first isosceles triangle, and ABD be the second
one. We want to proceed as in Hilbert’s theorem 17 (see Proposition 5.19 above).
Question. Why do the points A and B lie on different sides of CD? We need just to
use the Lemma 5.3 to SSS-congruence!
Answer. If points A and B would lie on the same side of line CD, the congruence
ACD ∼
= BCD would imply A = B, contradicting to the endpoints of a segment
being different.
169
Because A and B lie on different sides of line CD, the segment AB intersects the
line CD, say at point M . Now we have the symmetric kite ACBD with two congruent
triangles (left and right in the figure): ACD ∼
= BCD.
Question. For convenience, please repeat how this follows just as in Theorem 17 (see
Proposition 5.19 above).
Answer. Addition or subtraction of the base angles of the two isosceles triangles ABC
and ABD yields the congruent angles ∠CAD ∼
= ∠CBD—either using sums or differences of congruent base angles. The triangles ACD and BCD have congruent
angles at A and B, and two pairs of congruent adjacent sides AC ∼
= BC and AD ∼
= BD.
∼
Now SAS congruence implies ACD = BCD.
Of the three points M, C, D exactly one lies between the two others. We can assume
that C does not lie between D and M , this assumption can be achieved by possibly
interchanging the names C and D. Next we show that ACM ∼
= BCM . Indeed,
Figure 5.36: Get the perpendicular bisector via: (a) a convex kite, (b) a nonconvex kite
the triangles ACM and BCM have congruent angles at the common vertex C, and
two pairs of congruent adjacent sides AC ∼
= BC and M C ∼
= M C. Now we get by SAS
∼
congruence that ACM = BCM . Matching pieces are stressed in the drawing.
From the last triangle congruence we get ∠AM C ∼
= ∠BM C. Because M lies between
A and B, these are two congruent supplementary angles. Hence they are right angles.
Too, the triangle congruence implies AM ∼
= BM . Hence M is the midpoint of segment
AB. We have shown that the segment AB and the line CD intersect perpendicularly
at the midpoint of segment AB. Hence CD is the perpendicular bisector.
Remark. There are the possibilities of an (a) convex kite, or (b) a non convex kite.
(a): If the two isosceles ABC and ABD lie on different sides of AB, points C and
D are on different sides of AB, and it is clear that the segment CD intersects the
line AB.
170
(b): Too, it is possible to use two different isosceles triangles on the same side of AB.
To get the second isosceles triangle ABD, one transfers as base angles any
two congruent angles which are less than the base angles of the first triangle
ABC. The figure ACBD is still a symmetric—but non convex—kite. As proved
in Hilbert’s Theorem 17 (see Proposition 5.19 above), the two triangles left and
right in the figure are still congruent: ACD ∼
= BCD. The line CD still
intersects the segment AB, because points A and B are on different sides of CD,
but the two segments AB and CD do not intersect each other.
Construction 5.5 (Solution— the way one really wants it). Construct any isosceles
triangle over the given segment. Drop the perpendicular from its third vertex.
Figure 5.37: Construction of the perpendicular bisector—the natural way
Independent proof of validity. Let ABC be the isosceles triangle constructed at the
first step. In the next construction step, the base angle ∠BAC is reproduced along the
−→
ray AB to the other side of line AB, opposite to point C. Finally, one transfers segment
AC onto the newly produced ray, and gets the congruent segment AD ∼
= AC.
Question. Show the congruence
(up-down)
ABC ∼
= ABD
Answer. This follows by SAS congruence. Indeed, the two triangles have the common
side AB, the two sides AC ∼
= AD are congruent, and the angles ∠BAC and ∠BAD are
congruent, both by construction.
Hence the construction has produced two isosceles triangles with base AB, which
are congruent to each other. Next we show that the triangles
(left-right)
ACD ∼
= BCD
are congruent.
171
Figure 5.38: (up-down) ABC congruent ABD
(upleft-upright) ACM congruent M CB
(left-right) ACD congruent BCD
Question. Explain how this congruence is shown.
Answer. This follows by SAS congruence. All four sides AD ∼
= AC ∼
= BC ∼
= BD are
∼
congruent. Furthermore, the two triangles have congruent angles ∠CAD = ∠CBD, as
follows from the congruences ∠CAB ∼
= ∠DAB, and ∠ABC ∼
= ∠ABD of base angles of
the two isosceles triangles shown in (up-down)—and angle addition.
The line AB and the segment CD intersect, because C and D lie on different sides
of AB. I call the intersection point M . Finally we show that the triangles
(upleft-upright)
ACM ∼
= BCM
are congruent.
Question. Explain how this congruence is shown.
Answer. This follows by SAS congruence. Indeed, the two triangles have the common
side CM , the two sides AC ∼
= AB are congruent, and the angles ∠ACM and ∠BCM
are congruent because of congruence (left-right).
Hence the segments AM ∼
= BM are congruent— line CD bisects the segment AB.
Question. Why does M lies between A and B.
Answer. We know that the segments AM ∼
= BM are congruent—this is impossible for
a point on the line AB outside the segment AB.
Too, the angles ∠AM C ∼
= ∠BM C are congruent supplementary angles, and hence
right angles. Thus we have confirmed that CD is the perpendicular bisector of segment
AB, as to be shown.
Another attempt to prove validity, it does not yet work! As already shown in the construction for dropping a perpendicular, the line CD is perpendicular to the given line
AB. Let M be the intersection point of these two lines.
172
We need still to show that M is the midpoint of segment AB. To this end, we prove
that the triangles AM C ∼
= BM C are congruent. This can be shown only using
SAA-congruence from Proposition 5.45 below!
By the converse isosceles triangle Proposition, we know that AC ∼
= BC. Because
of congruence of vertical angles, we know that ∠BM C ∼
∠AM
D,
the
latter angle was
=
already shown to be a right angle. Hence, all put together, we get ∠BM C ∼
= ∠AM D ∼
=
∼
∼
R = ∠AM C as expected. Finally, we have the congruence base angles ∠BAC = ∠ABC
by construction.
Hence via SAA-congruence from Proposition 5.45 below, we look forward to establishing the triangle congruence AM C ∼
= BM C, and hence confirming AM ∼
=
M B.
Question. Why does this proof not work, at this point of the development?
Answer.
5.7
The exterior angle theorem and its consequences
Recall that an exterior angle of a triangle is the supplement of an interior angle.
Proposition 5.34 (The Exterior Angle Theorem). [Euclid I.16. Theorem 22 in
Hilbert] The exterior angle of a triangle is greater than both nonadjacent interior angles.
−→
Proof. For the given ABC, we can choose point D on the ray opposite to AB, such
that AD ∼
= CB. We compare the two nonadjacent interior angles γ = ∠ACB and
β = ∠ABC to the exterior angle δ = ∠CAD. As a first step, we show
Lemma 5.4. The exterior angle δ = ∠CAD is not congruent to the interior angle
∠ACB = γ.
Proof of Lemma‘5.4. Assume towards a contradiction that δ ∼
= γ.
The supplementary angle of δ is α = ∠CAB. The supplementary angle of γ is
−−→
∠ACE, with a point E on the ray opposite to CB. Supplementary angles of congruent
angles are congruent (Theorem 14 in Hilbert, see Proposition 5.14 above). Hence
?1
∠CAB ∼
= ∠ACE
On the other hand, we use SAS-congruence for the two triangles ABC and A B C with
A := C , B := D , C := A
Those two triangles would be congruent by SAS congruence. Indeed the angles at
C and C are congruent by our assumption γ = ∠ACB ∼
= ∠CAD = δ. Too, the
adjacent sides are pairwise congruent because of CB ∼
= AD = C B by construction,
173
Figure 5.39: The impossible situation of a congruent exterior angle
and CA ∼
= AC = C A . From Axiom III.5, we conclude that the corresponding angles
at A and A = C are congruent:
?2
∠CAB ∼
= ∠C A B = ∠ACD
We have shown that angle ∠CAB is congruent to both (?1) ∠ACE and (?2) ∠ACD.
Now we use the uniqueness of angle transfer. Transferring angle ∠CAB with one
−→
side CA, into the half plane not containing B, yields as second side of the angle once
−−→
−−→
the ray CE and the second time CD. Hence these two rays are the same.
Hence points E and D both lie on the line la := BC. In other words, the four points
B, C, D, E all lie on one line. Because B and D lie on opposite rays with vertex A,
these two points are different: B = D. Note that just this simple remark is not true in
elliptic geometry!
Because a line is uniquely specified by two of its points, this implies la = BC = BD.
We have just seen that point C lies on this line. By construction, point A lies on the
line BD, too. Hence all the points A, B, C, D, E lie on the same line. This contradicts
the definition of a triangle. By definition of a triangle, its three vertices do not lie on
one line. This contradiction confirms the original claim δ γ.
Remark. To get a reminiscence to point symmetry about the midpoint M of AC, we can
choose AB ∼
= CE. With the choice AB ∼
= CE, we get even E = D, but still D = B.
Remark. In spherical geometry, the figure constructed does exist. We do not get a
contradiction, the conclusion is just that the two points B and D are antipodes, and
the vertices A and C lie on two different lines BAD and BCD through these two
antipodes. Too, the sum of the segments is congruent going both ways from B to D.
The triangles ABC ∼
= A B C = CDA are congruent, by SAS congruence (given
in Proposition 5.10 above). Hence BC ∼
= DA and CD ∼
= AB, and hence the sums
∼
BC + CD = DA + AB are congruent.
The next Lemma rules out the possibility δ < γ.
174
Figure 5.40: Two ways from B to D
Lemma 5.5. The exterior angle δ = ∠CAD is not smaller that the interior angle
∠ACB = γ.
Proof of Lemma 5.5. Suppose towards a contradiction that δ < γ. We transfer the
Figure 5.41: The impossible situation of a smaller exterior angle
−→
exterior angle δ = ∠CAD with one side CA, into the half plane containing B. The
−−→
second side of the angle we get is a ray CB inside the angle ∠ACB. By the Crossbar
Theorem, it meets the segment AB in a point B .
We can now apply the first Lemma 5.4 to AB C. This smaller triangle would have
exterior angle δ congruent to the interior angle γ = ∠ACB . This is impossible by
Lemma 5.4.
Thus we have both ruled out the possibility that δ = γ, or that δ < γ. By Proposition 5.26, all angles are comparable. Hence there remains only the possibility that
δ > γ.
Lemma 5.6. The exterior angle δ = ∠CAD is greater than the interior angle
β = ∠ABC.
Proof of Lemma 5.6. We compare angle β = ∠ABC to the exterior angle δ = ∠CAD.
−→
Choose any point F on the ray opposite to AC. The vertical angles δ = ∠CAD ∼
=
∠F AB = δ are congruent by Euclid I.15. Now we are back to the case already covered,
because the interior angle β = ∠ABC and exterior angle ∠F AB lie on opposite sides
of triangle side AB, which is also part of one side of each of these angles. By Lemma5.4
and 5.5, we conclude ∠F AB > ∠ABC and hence δ > β.
175
Figure 5.42: Comparing the other nonadjacent angle
Remark. Here is an other way to prove Lemma 5.6: Define A3 B3 C3 by setting A3 :=
A, B3 := C, C3 := B. Now we can apply Lemma5.4 and 5.5 to this new triangle and
get δ = δ3 > γ3 = β. By congruence of the vertical angles δ = ∠CAD ∼
= ∠F AB = δ we get δ > β.
Thus the proof of the exterior angle theorem is finished.
Proposition 5.35. [Euclid I.17.] The sum of any two interior angles of a triangle is
less than two right angles.
Proposition 5.36 (Immediate consequences of the exterior angle theorem).
(i) Every triangle can have at most one right or obtuse angle.
(ii) The base angles of an isosceles triangle are acute.
(iii) The foot point of a perpendicular is unique.
(iv) Given a line l and a point O not on l. At most two points of l have the same
distance from O.
(v) A circle and a line can intersect in at most two points.
Proof. (i): Suppose angle α of ABC is right or obtuse: α ≥ R. By Proposition 5.31,
its supplement is right or acute: S(α) ≤ R. But the supplement is an exterior
angle: δ = S(α). By the exterior angle theorem, the two other nonadjacent
interior angles of the triangle are less than that exterior angle. Hence they are
acute: β, γ < δ = S(α) ≤ R, and hence β, γ < R.
(ii): The two base angles being congruent, by (i), they cannot be both right or obtuse.
176
(iii): Given a line l and a point O not on l. Suppose there are two perpendiculars, with
foot points F and G. The OF G would have two right angles, contradicting (i).
(iv): Given a line l and a point O not on l. Suppose towards a contradiction there
Figure 5.43: No three points have the same distance from a line
are three points A, B, C on the line l such that OA ∼
= OB ∼
= OC. Of any three
points, one lies between the two others (Theorem 4 of Hilbert). We may suppose
by renaming that A ∗ B ∗ C.
The base angle α of triangle OBC is an exterior angle of the other triangle
OAB. Hence the exterior angle theorem yields γ > α. But the role of the two
triangles can be switched: The base angle γ of triangle OAB is an exterior angle
of the other triangle OBC. Hence the exterior angle theorem yields α < γ. Both
inequalities together are impossible. This contradiction implies there can exist at
most two points on a line which have congruent distances from point O.
(v) For the case that the center O of the circle does not lie on the line, the statement
is confirmed by item (iv) above.
For the case that the center O of the circle lies on the line, the statement follows
from line separation and the uniqueness of segment transfer stated in Proposition 5.2.
Definition 5.11 (Alternate interior angles or z-angles). Let a transversal t intersect two lines a and b at the points A and B. A pair of alternate interior angles or simply
z-angles are two angles with vertices A and B, lying on different sides of the transversal.
They have as one pair of sides lies on the transversal and contains the segment AB, the
remaining two sides are rays lying on a and b.
Proposition 5.37 (Congruent z-angles imply parallels). [Euclid I.27] If two lines
form congruent z-angles with a transversal, they are parallel.
Proof. This is an immediate consequence of the exterior angle theorem. We argue by
contradiction. Suppose the two lines would intersect in point C. One of the z-angles
177
is an exterior angle of triangle ABC, the other one is a nonadjacent interior angle.
Lemma 5.4 does imply that the two z-angles are not congruent— contradicting the
assumption from above. Hence the two lines a and b cannot intersect.
Figure 5.44: A pair of z-angles.
Remark. Referring to the pair of z-angles shown in the figure on page 178 the Proposition 5.37 tells that
α∼
=β⇒ab
This is equivalent to its contrapositive
a∦b⇒α∼
= β
which relates even more directly to the exterior angle theorem.
As a consequence of Euclid I.27, the existence of a parallel can be proved in neutral
geometry. One parallel to l through point P is conveniently constructed as ”double
perpendicular”.
Proposition 5.38 (Existence of a parallel). For every line l and for every point P
lying not on l, there exists at least one parallel m to l through point P .
Proof. Given is line l and a point P not on l. One drops the perpendicular from point P
onto line l and denotes the foot point by F . Next, one erects at point P the perpendicular
to line P F . Thus, one gets a line, which we call m. Because the two lines l and m form
←→
congruent z-angles with the transversal P F , Euclid I.27 or I.28 imply that l and m are
parallel. This leaves open the question whether or not m is the unique parallel to line l
through point P .
178
Recall that the Euclidean Parallel Postulate 2.2 postulates both existence and uniqueness of the parallel to a given line through a given point. On the other hand, Hilbert’s
Parallel Postulate for plane geometry 2.3 postulates only the uniqueness of the parallel
to a given line through a given point. Because of Proposition 5.38, Euclid’s and Hilbert’s
parallel postulate turn out to be equivalent for any Hilbert plane. In other word, we
can state the definition equivalent to definition 1.2:
Definition 5.12 (Pythagorean plane). A Pythagorean plane is a Hilbert plane for
which the Euclidean parallel postulate holds.
Problem 5.6. Given is a convex quadrilateral with two pairs of congruent opposite
sides. Since the quadrilateral is convex, each diagonal separates the quadrilateral into
two triangles in opposite half planes of their common side. Prove in neutral geometry
that the two pairs of opposite sides are parallel. Provide a drawing.
Figure 5.45: A parallelogram in neutral geometry.
Answer. Given is the quadrilateral ACBD with two pairs of congruent opposite sides
AC ∼
= BD and BC ∼
= AD. We draw the diagonal AB and get the congruent triangles
ABC ∼
BAD,
as
one confirms by SSS congruence. Hence the angles β = ∠ABC
=
and α = ∠BAD are congruent. We see that the diagonal AB transverses the lines of
the opposite sides DA and BC with congruent z-angles. By Euclid I.27, we conclude
that the opposite sides are parallel.
Remark. Similar to the exterior angle theorem, these are facts of neutral geometry. They
are is valid both in Euclidean and hyperbolic geometry. Note that Euclid I.2 through
I.28 are theorems that hold for every Hilbert plane. Indeed Euclid I.29 is the first
theorem in Euclid’s elements that uses the Euclidean parallel postulate.
Warning. The converse of Euclid I.27 is does not hold in hyperbolic geometry. Indeed,
in hyperbolic geometry, parallels can form non-congruent z-angles with a transversal.
179
Only in Euclidean geometry is it true that a b ⇒ α ∼
= β. In more common
language:
Only in Euclidean geometry can parallel lines be used as a means
for the transport of an angle.
Problem 5.7 (Addendum the the extended ASA-Congruence Theorem). In
the situation of the extended ASA-congruence theorem, what happens in case that A B <
AB or A B > AB?
Question. Using the exterior angle theorem and Pasch’s axiom, get some results for the
first case.
Answer. In the case A B < AB, one can conclude that the two rays rA
and rB
still do
intersect.
Detailed proof. I show that the two rays do intersect. One transfers segment AB to the
−−→
ray A B and gets a segment A B2 ∼
= AB with A ∗ B ∗ B2 .
We apply the extended ASA theorem to ABC and segment A B2 , and get a triangle
A B2 C2 ∼
= ABC.
Now apply Pasch’s axiom to triangle AB2 C2 and line l on the ray newly produced
rB
. This line intersects the triangle side A B2 in B , by construction. Hence Pasch’s
axiom tells that line l intersects a second side of AB2 C2 , too, or goes through point
C2 .
Line l does not go through point C2 . Otherwise, one would get a contradiction to
the exterior angle theorem.
Question. For which triangle do you get a this contradiction?
Answer. B B2 C would have both the interior angle β at vertex B2 , and the exterior
angle β at vertex B .
Line l does not intersect line B2 C2 . Again, one would get a contradiction to the
exterior angle theorem. Alternatively, one can say that the two lines l and B C are
parallel, because they form congruent z-angles with line A B .
Hence Pasch’s axiom implies that line l intersects the third side of AB2 C2 , which
is segment A C2 . We call the intersection point C . Thus we have produced a A B C ,
the angles of which at vertices A and B are congruent to the corresponding angles of
ABC.
Remark. In Euclidean geometry, the two rays rA
and rB
always intersect, no matter
∼
whether A B < AB, A B = AB, or A B > AB. In all three cases, the A B C is
similar to ABC.
Remark. Here is what happens in hyperbolic geometry: In hyperbolic geometry, similar
triangles are always congruent, this implies that A B C and ABC are not similar
if either A B < AB or A B > AB!
180
Figure 5.46: The extended ASA congruence once more
In the case just considered, A B < AB and γ > γ. In the opposite case that
A B > AB, the second triangle A B C has either a different angle γ < γ, or does
not exist at all. The last possibilities occurs because the two rays rA
and rB
do not
intersect at all. This happens always, once the segment A B is long enough.
Proposition 5.39 (Comparison of sides implies comparison of angles). [Euclid
I.18, Theorem 23 of Hilbert] In any triangle, across the longer side lies the greater angle.
Proof. In ABC, we assume for sides AB and BC that c = AB > BC = a. The issue
is to compare the angles α = ∠CAB and γ = ∠ACB across these two sides.
We transfer the shorter side BC at the common vertex B onto the longer side. Thus
one gets a segment BD ∼
= BC, with point D between B and A. Because the BCD is
isosceles, it has two congruent base angles
δ = ∠CDB ∼
= ∠DCB
Because B ∗ D ∗ A, we get by angle comparison at vertex C
δ = ∠DCB < γ = ∠ACB
Now we use the exterior angle theorem for ACD. Hence
α = ∠CAB < δ = ∠CDB
By transitivity, these three equations together imply that α < γ. Hence the angle α
across the smaller side CB is smaller than the angle γ lying across the greater side AB.
In short, we have shown that c > a ⇒ γ > α.
Proposition 5.40 (Comparison of angles implies comparison of sides). [Euclid
I.19] In any triangle, across the greater angle lies the longer side.
Proof. In ABC, we assume for two angles that α = ∠CAB < γ = ∠ACB. The issue
is to compare the two sides BC and AB lying across the angles and show a = BC <
AB = c. As shown in Proposition 5.6, any two segments are comparable.
181
Figure 5.47: Across the longer side lies the greater angle
Question. Please repeat the reasoning for convenience.
−→
Answer. We transfer segment BC along the ray BA, and get a segment BD ∼
= BC. By
Theorem 4 in Hilbert, of the three points A, B, D on a line AB, exactly one lies between
the two others. This leads to the following three cases: Either
(i) A ∗ D ∗ B and a < c. or (ii) A = D and a = c. or (i) D ∗ A ∗ B and a > c.
We can now rule out cases (ii) and (iii).
In case (ii), the ABC is isosceles, and Euclid I.5 would imply α = γ, contrary to
the hypothesis about these two angles.
In case (iii), Euclid I.18 or Theorem 23 of Hilbert, would imply α > γ, contrary to
the hypothesis.
Hence only case (i) is left, and a < c, as to be shown. In short, we have shown that
γ > α ⇒ c > a.
Question. Explain how Euclid I.18 and Euclid I.19 are logically related.
Answer.
Euclid I.18 in shorthand: c > a ⇒ γ > α.
Euclid I.19 in shorthand: γ > α ⇒ c > a.
Euclid I.19 is the converse of Euclid I.18.
Question. Does Euclid I.19 follow from Euclid I.18 by pure logic? Why not?
Answer. No, the converse does not follow purely by logic.
Question. Which fact does the proof above work nevertheless?
Answer. Because any two segments are comparable, we get the converse, nevertheless.
Corollary 8. A triangle with two congruent angles is isosceles.
Proof. Assume α ∼
= γ for the given triangle. Since c > a ⇒ γ > α and c < a ⇒ γ < α,
and any two segments are comparable, only a ∼
= c is possible.
182
Proposition 5.41. The hypothenuse is the longest side of a right triangle.
Proof. By Proposition 5.36(i), a triangle can have at most one right or obtuse angle.
Hence a right triangle ABC with the right angle at vertex C has acute angles at the
vertices A and B. As shown in Proposition 5.40 (Euclid I.19), comparison of angles of
a triangle implies comparison of its sides. Hence the two legs across of the acute angles
A or B are shorter than the hypothenuse across the right angle.
Proposition 5.42 (The foot point has the shortest distance). Given is a line and
a point O not on the line. The foot point F of a perpendicular from O to the line is the
unique among all points of the line, which has the shortest distance.
Figure 5.48: The foot point has the shortest distance
Proof. Call the given line l and let O be the given point not on l. Take any point A = F
on the line l. The AF O has a right angle at vertex F . The hypothenuse AO across
to F is the longest side.
Remark. Because all other point on l except the foot point have strictly greater distance
from O, we can once more conclude that the foot point of the perpendicular is unique.
Proposition 5.43 (The Triangle Inequality). [Euclid I.20] In any triangle, the sum
of two sides is greater than the third side.
Proof. In ABC, we compare the sum BA + AC to the third side BC. According to
Definition 5.2 the sum BA + AC is obtained by transfer of the segment AC to the ray
−→
opposite to AB. We get segment DA ∼
= AC and an isosceles DAC. Its two congruent
base angles are
δ = ∠CDA ∼
= ∠DCA
Because point A lies between D and B, angle comparison at vertex C yields
δ∼
= ∠DCA < ∠DCB = η
183
Figure 5.49: The triangle inequality
Now we use Euclid I.19 for the larger DCB. Hence the side BC, across the smaller
angle δ is smaller than the side DB lying across the greater angle η: DB > BC. For
the original ABC, this shows that indeed CA + AB > CB.
Corollary 9. For any three points A, B, C we have the inequality
AC ≤ AB + BC
and AC ∼
= AB + BC holds if and only if point B lies between A and C or is equal to
one of them.
Proposition 5.44. The sum of the segments of any polygon is equal to or longer than
the distance of its first and last vertex.
The sum of the polygon segments A0 A1 , A1 A2 , A2 A3 , . . . , An−1 An is equal to the distance of its first vertex A0 to its last vertex An if and only if all vertices lie between them
and occur in the order A0 ∗ A1 ∗ A2 ∗ A3 ∗ · · · ∗ An .
Proof. We proceed by induction on the number of segments of the polygon
A0 A1 , A1 A2 , A2 A3 , . . . , An−1 An . The assertion holds for a polygon A0 A1 , A1 A2 with
two segments since A0 A2 ≤ A0 A1 + A1 A2 according to the triangle inequality, and its
corollary.
For the induction step, we proceed from n to n + 1 segments. We know that
A0 An+1 ≤ A0 An + An An+1
by the triangle inequality and
A0 An ≤ A0 A1 + A1 A2 + · · · + An−1 An
by the induction assumption. The associativity of the segment addition has already
been stated as Proposition 5.7. Hence
A0 An+1 ≤ A0 An + An An+1 ≤ (A0 A1 + A1 A2 + · · · + An−1 An ) + An An+1
∼
= A0 A1 + A1 A2 + · · · + An−1 An + An An+1
as to be shown.
184
Proposition 5.45 (SAA-Congruence Theorem). [Theorem 25 in Hilbert] Assume
two triangles have a pair of congruent sides, one pair of congruent angles across these
sides, and a second pair of congruent angles adjacent to these sides. Then the two
triangles are congruent.
Figure 5.50: SAA congruence
Proof. Let the triangles be ABC and A B C , and assume that AB ∼
= A B , ∠BAC ∼
=
−−
→
∼
∠B A C and ∠ACB = ∠A C B . We choose a point C on the ray A C such that
AC ∼
= A C . By axiom III.5
γ = ∠ACB ∼
= ∠A C B Indeed, the SAS congruence even implies
(*)
ABC ∼
= A B C On the other hand, by assumption
γ = ∠ACB ∼
= ∠A C B If C = C , an exterior angle of C C B would be congruent to a nonadjacent interior
angle, which is impossible by the exterior angle theorem. Hence we can conclude that
C = C . Because of (*), this implies ABC ∼
= A B C , as to be shown.
Question. How does this theorem differ from the ASA congruence (Hilbert’s Theorem
13, and Proposition 5.12 above)?
Answer. Of the two pairs of angles that are given (or compared), one pair lies across
the pair of given sides.
Second proof of the hypothenuse-leg theorem 5.30. Take two right ABC and A B C with b = b , c = c and γ = γ = R. One transfers segment CB and segment C B , both
185
Figure 5.51: The hypothenuse leg theorem
−−→
on the ray opposite to C B , and get new segments C D0 ∼
= CB as well as C E ∼
= C B.
From the construction and SAS congruence, we conclude
ABC ∼
= A D0 C ,
(1)
A B C ∼
= A E C Hence especially
(2)
A B ∼
= A D0 ∼
= A E By Proposition 5.36(iv), at most two points can have the same distance from a line.
Hence not all three points B , D0 , E can be different. The only possibility left is D0 = B ,
because they both lie on the opposite side of A C than B .
Question. For completeness, explain once more. Assume that D0 = E towards a contradiction. Take the case that B ∗ D0 ∗ E shown in the drawing. (The other cases can
be dealt with similarly.)
Independent answer. There are two isosceles B A D0 and DA E .
Question. What would happen with their base angles at vertex D0 ?
Answer. The two base angles would be supplementary.
Question. Why is this impossible?
Answer. This is impossible, because the exterior angle theorem would imply that each
one of them is larger than the other one.
This contradiction leave only the possibility that D0 = E .
Now the required congruence ABC ∼
= A B C follows from D0 = E and (1).
Definition 5.13 (The angular bisector). The ray in the interior of an angle which
bisects the angle into two congruent angles is called the angular bisector.
186
Proposition 5.46 (Uniqueness). The angular bisector is unique.
−−→
−−→
Proof. Suppose both rays BG and BG bisect angle ∠BAC. The angles α := ∠ABG and
α := ∠ABG are comparable, because all angles are comparable by Proposition 5.26.
Suppose towards a contradiction that α < α . As specified in Proposition 5.29, one
can add inequalities of angles. Hence we conclude that
(?)
∠ABC = ∠ABG + ∠GBC ∼
= ∠ABC
= α + α < α + α ∼
= ∠ABG + ∠G BC ∼
which is impossible. Similarly, the case α > α can be ruled out. Hence α ∼
= α , and
−−→ −−→
hence by uniqueness of angle transfer BG = BG , as to be shown.
Problem 5.8. Given is any angle ∠BAC. Construct the angular bisector.
Construction 5.6. We choose the segments on its sides to be congruent, thus assuming
−−→
AB ∼
= AC. Draw the line BC, and transfer the base angle ∠ABC to the ray BC, on
the side of line BC opposite to vertex A. On the new ray, we transfer segment AB to
−−→
get the new segment BD ∼
= BA. The ray AD is the bisector of the given angle ∠BAC.
Figure 5.52: The angular bisector
Question. Reformulate the description of this construction precisely, and as short as
possible.
Answer. One transfers two congruent segments AB and AC onto the two sides of the
angle, both starting from the vertex A of the angle. The perpendicular, dropped from
the vertex A onto the segment BC, is the angular bisector.
Proof of validity. By assumption, the three points A, B, C do not lie on a line. By
construction, points A and D lie on different sides of line BC. Hence the segment AD
intersects line BC, say at point M . In step (1) and (2), we get three congruent triangles.
Step (1): We confirm that AM B ∼
= DM B. The matching pieces used for the proof
are stressed.
187
Figure 5.53: The first pair of congruent triangles
Answer. Indeed, by construction, ∠ABC ∼
= ∠DBC. Hence ∠ABM ∼
= ∠DBM . (It
−−→
does not matter whether M lies on the ray BC or the opposite ray.) Too, we have a
pair of congruent adjacent sides: Indeed BD ∼
= BA by construction, and BM ∼
= BM .
∼
Now SAS congruence implies AM B = DM B.
Question. Explain carefully why ∠AM B is a right angle.
Answer. Because of AM B ∼
= DM B, we get ∠AM B ∼
= ∠DM B. Because point M
lies between A and D, these are two supplementary angles. Hence they are right angles.
Step (2): Finally, we confirm that AM B ∼
= AM C. Again the pieces needed to
match for this theorem are stressed in the drawing.
Figure 5.54: The second pair of congruent triangles
Answer. Indeed, I use the hypothenuse-leg theorem. ∠AM B ∼
= ∠AM C ∼
= R, because
a right angle is congruent to its supplement. (Again, it does not matter whether M lies
−−→
on the ray BC or the opposite ray.) Too, we have a pair of congruent sides: Indeed
AB ∼
= AC by construction, and AM ∼
= AM .
From the triangle congruence AM B ∼
= AM C, we get ∠M AB ∼
= ∠M AC, and
∼
M B = M C. Since point M lies on the line BC, the last congruence shows that M lies
188
−−→ −−→
between B and C, too. Hence ray AM = AD lies inside the given ∠BAC and bisects
this angle.
Proposition 5.47 (Existence of the angular bisector). For any angle ∠BAC, there
−−→
exists a ray AD inside the given angle such that ∠DAB ∼
= ∠DAC.
Proposition 5.48 (The Hinge Theorem). [Euclid I.24] Increasing the angle between
two constant sides increases the opposite side of a triangle.
Figure 5.55: The Hinge Theorem
Corollary 10 states the equivalence we get by taking from [Euclid I.24] and [Euclid I.25]
together:
Corollary 10. Given are two triangles with two pairs of congruent sides. In the first
triangle, the angle between them is smaller, congruent, or greater than in the second one
if and only if the opposite side is smaller, congruent, or greater in the first triangle.
Proof. Given are ABC and A B C with a = a and c = c . Assuming β < β , we
have to check whether b < b . One can assume that A = A , B = B , and put the two
points C and C lie on the same side of line AB. By the hypothesis β < β , the two
points A and C lie on different sides of line BC. Hence this line intersects the segment
AC .
−−→
Indeed, as stated by the Crossbar theorem, the ray BC intersects the segment AC .
Let D be the intersection point. We now distinguish three cases:
(a) B ∗ D ∗ C. This case occurs if the triangle side BC to be rotated is long enough.
(b) C = D.
189
Figure 5.56: The Hinge Theorem: turning a long side
Figure 5.57: The Hinge Theorem: the borderline case
(c) B ∗ C ∗ D. In this case the triangle side BC to be rotated is rather short.
In the border line case (b), we see directly that β < β implies A ∗ C ∗ C and hence
AC < AC , as to be shown. In the two other cases, we apply Euclid I.19 to triangle
ACC . Thus it is enough to show that this triangle has a larger angle ε = ∠ACC at
vertex C than the angle ε = ∠AC C at vertex C . The BCC is isosceles. Hence, by
Euclid I.5, it has two congruent base angles, which we denote by ϕ.
In case (a), we proceed as follows: Because B and C lie on opposite sides of AC ,
comparison of angles at vertex C yields
(1)
ε < ϕ
190
Figure 5.58: The Hinge Theorem: turning a short side
And because A and C lie on opposite sides of line BC, comparison of angles at vertex
C yields
(2)
ϕ<ε
Because of transitivity, (1)(2) together imply ε < ε and hence AC > AC.
In case (c), we show that ε is an obtuse, and ε is an acute angle. Because A and C lie
on opposite sides of line BCD, we get
∠C CD < ∠C CA = ε
(3)
but ∠C CD is an exterior angle of the isosceles BCC . Because the base angles of
an isosceles triangle are acute, its supplement ∠C CD is obtuse. Hence by (3), ε is
obtuse, too. Since the triangle ACC can have at most one right or obtuse angle, the
angle ε is acute. Now ε < R < ε implies again ε < ε, and hence Euclid I.19 yields
AC > AC.
Proposition 5.49 (The Midpoint of a Segment). [Theorem 26 in Hilbert] Any
segment has a midpoint.
Construction 5.7. Given is a segment AB. Transfer congruent angles with the endpoints of the given segment AB as vertices, on different sides of AB. Next we transfer
congruent segments AC ∼
= BD onto the newly produced legs of these two angles. Finally,
←→
←→
lines AB and CD intersect at the midpoint M .
(a) Here is a drawing.
(b) Explain why lines AB and CD intersect.
Answer. By construction, points C and D lie on different sides of line AB. Hence,
by the plane separation theorem, line AB and segment CD intersect.
191
Figure 5.59: Hilbert’s construction of the midpoint
Let M be the intersection point. From the plane separation theorem, too, it follows
that the intersection point M lies between C and D. But it turns out to be harder
to see why M lies between A and B!
(c) Show that M = A is impossible.
Figure 5.60: M = A is impossible
−→
Answer. In that case line l = AC would go through point D. The ray AC would
be an extension of side AD of the triangle ABD. This triangle would have the
interior angle ∠ABD congruent to the exterior angle ∠CAB, contradicting the
exterior angle theorem. (Remember: an exterior angle is always greater than a
nonadjacent interior angle.)
192
(d) Show that M ∗ A ∗ B is impossible.
Figure 5.61: Point A lying between M and B is impossible
Answer 1. We use Pasch’s axiom for M BD and line l = CA. Which conclusion
do you get?
Answer. The line CA enters M BD on the side M B. By Pasch’s axiom, this
line either (i) intersects side DB, or (ii) goes through point D, or (iii) intersects
side M D.
In all three cases, we derive a contradiction:
Case (i): suppose line l intersects segment DB, say at point G. The ABG
would have the interior ∠ABG congruent to the exterior ∠CAB. This contradicts the exterior angle theorem, which tells an exterior angle is always
greater than a nonadjacent interior angle.
Case (ii): suppose line l goes through point D. The ABD would have the
interior ∠ABD congruent to the exterior ∠CAB. This contradicts the exterior angle theorem, which tells an exterior angle is always greater than a
nonadjacent interior angle.
Case (iii): suppose line l intersects segment M D, say at point F . Points C = F
are different, because they lie on different sides of AB. The lines AC and
M D intersect both in point C and in point F ,. Because this are two different
points, they determine a line uniquely. Hence all five points C, M, D, A, F lie
on one line. Hence we are back to the case M = A, ruled out earlier.
Answer following Hilbert. We apply the exterior angle theorem twice. Here is a
sketchy drawing for that impossibility. 26 In triangle ABD, the interior angle
26
The drawing, too, occurs in the millenium edition of ”Grundlagen der Geometrie”, page 26.
193
Figure 5.62: Point A lying between M and B is impossible
at vertex B is β = ∠DBM , which is smaller than the exterior angle at vertex
ε = ∠BM C. In triangle AM C, the angle ε from above is interior angle at
vertex M , and hence smaller than the exterior angle α = ∠BAC at vertex A.
Now transitivity yields β < ε < α. On the other hand, the angles ∠DBA = β
and α = ∠CAB are congruent by construction. This contradiction rules out the
case M ∗ A ∗ B.
(e) Now we know that M lies between A and B, we finally can prove that M is the
midpoint.
Question. Which congruence theorem is used for which triangles?
Answer. One uses SAA congruence for AM C and BM D. Indeed, the angles at
Figure 5.63: Apply the SAA congruence
194
A and B are congruent by construction, and the angles at vertex M are congruent
vertical angles. (Both statements are only true because M lies between A and B!)
The sides AC and BD opposite to those angles are congruent by construction.
Hence the two triangles are congruent, and especially AM ∼
= MB
Figure 5.64: The generic situation for Proposition 5.50, for which we prove: Two segments
CX and DY on different sides of XY are congruent if and only if midpoint M of segment CD
lies on line l. The third figure shows the case with both conditions true.
Problem 5.9. Given is a convex quadrilateral with two pairs of congruent opposite
sides. Prove that the diagonals bisect each other.
Use the result of problem 5.6 and assume Hilbert’s construction of the midpoint of
a segment is valid. Provide a drawing.
Figure 5.65: The diagonals of a parallelogram bisect each other.
Answer. Since the quadrilateral is convex, the opposite vertices C and D lie on different
sides of the diagonal AB.
195
We have proved in problem 5.6 that the angles β = ∠ABC and α = ∠BAD are
congruent. Hence the figure repeats exactly Hilbert’s construction of the midpoint of
segment AB. Hence the diagonal segments intersect in the midpoint M of diagonal AB.
Similarly, we prove that M is the midpoint of the other diagonal CD, too.
Proposition 5.50. The midpoint of a segment lies on a given line l if and only if the
two endpoints of a segment have the same distance to the line, and lie on the opposite
sides of it.
Proof. Assume that the two endpoints C and D of the given segment lie on opposite
sides of l. Furthermore, assume CX ∼
= DY are the congruent segments to the foot
points X and Y .
Let Q be the intersection point of line l and segment CD, which exists by plane
separation. In the special case X = Y , we are ready immediately. Otherwise, we need
to see why Q lies between X and Y ! This is done is the same way as in Hilbert’s
construction 5.7 of the midpoint, which is now applied to the segment AB = XY .
Finally, one obtains the triangle congruence
CQX ∼
= DQY
(oneflier)
via SAA congruence, using the right angles at X and Y , vertical angles at vertex M ,
and the congruent segments CX ∼
= DY . Hence CQ ∼
= DQ, confirming that Q = M is
the midpoint of segment CD.
Conversely, assume that the midpoint M of segment CD lies on the line l. In
the special case X = Y we are ready immediately. Otherwise, we need to confirm, once
more, why M lies between X and Y ! Again one needs to use the exterior angle theorem.
Finally, one obtains the triangle congruence
CM X ∼
= DM Y
(oneflier)
via SAA congruence, using the right angles at X and Y , vertical angles at vertex M ,
and the congruent segments CX ∼
= DY , as to be shown.
5.8
SSA congruence
Next we study the possibilities and difficulties with SSA congruence. Thus the matching
pieces of the two given triangles are two pairs of sides, and one pair of angles opposite
to one of these sides. I follow Euler’s conventional notation: in triangle ABC, let
a = BC, b = AC, and c = AB be the sides and α := ∠BAC, β := ∠ABC, and
γ := ∠ACB be the angles.
Problem 5.10. Investigate an example Use straightedge and compass to construct
in Euclidian geometry all triangles (up to congruence) with γ = ∠ACB = 30◦ , side
AC = 10 units and side AB given as below. How many non-congruent solutions (none,
196
one, two?) do you get in each case? How many of them are acute, right or obtuse
triangles? Measure and report the angle β = ∠ABC for all your solutions. Make clear
by the drawings what happens in all cases, especially how many acute, right, and obtuse
triangles you get as solutions.
Hint: It is convenient to construct all triangles with one common side AC. Check
carefully to find the obtuse angles! (a) AB = 4 units.
Figure 5.66: A Euclidean example for SSA triangle construction
Answer (a). One has to begin the construction by drawing a segment AC of length 10,
←−
and a ray r with vertex C, forming an angle of 30◦ with CA. The point B has to lie on
the ray r, as well as on a circle of radius AB around A. In case (a), the circle does not
intersect this ray, hence there is no solution.
(b) AB = 5 units.
Answer. In case (b), the circle just touches the ray r, hence there is one solution, with
a right angle at B.
(c) AB = 5.5 units.
Answer. In case (c), the circle intersect the ray r in two points, hence there are two non
congruent solutions. At vertices B and B , I measure the angles about β = 67◦ and
β = 113◦ . One solution is an acute, the other an obtuse triangle.
(d) AB = 6 units.
197
Answer. Again in case (d), the circle intersects the ray r in two points, and there are
two non congruent solutions. At vertices B and B , I measure the angles about β = 57◦
and β = 123◦ . Both solutions are obtuse triangles. Indeed, the first solution has the
obtuse angle α = 180◦ − β − γ = 93◦ .
Proposition 5.51 (SSA Matching Proposition). Given are two triangles. Assume
that two sides of the first triangle are pairwise congruent to two sides of the second
triangle, and that the angles across to one of these pairs are congruent.
Then either the two triangles are congruent, or the angles across the other pair of
congruent sides add up to two right angles.
Proof. It is assumed that two sides and the angle across one of these sides of ABC
can be matched to congruent pieces of A B C :
(SSA)
c = AB ∼
= A B = c , b = AC ∼
= A C = b , γ = ∠ACB ∼
= ∠A C B = γ We have to show that either (a) or (b) holds:
(a) ABC ∼
= A B C .
(b) The two angles across the second matched side β = ∠ABC and β = ∠A B C are
(congruent to) supplementary angles. One of them is acute and the other one is
obtuse.
We begin by reducing the problem to the special case that A = A , C = C and that
B and B = D lie on the same side of AC. We use the SAS congruence to construct a
triangle ADC, such that A B C ∼
= ADC, with points B and D on the same side
of AC. The drawing below shows that procedure.
Figure 5.67: Matching two triangles with SSA
Question. Explain how you get the triangle ADC congruent to A B C .
198
−−→
Answer. I just transfer segment C B onto the the ray CB and get C B ∼
= CD. The
∼
congruence A B C = ADC, follows from SAS, given in Theorem 12 of Hilbert (see
Proposition 5.10 above).
Assume that congruence (a) does not hold. We have to prove that case (b) occurs.
We know that α = α , since otherwise the SAS-congruence theorem implies ABC ∼
=
A B C , which again would be case (a) just ruled out. Without loss of generality, we
can assume α < α. Since ∠DAC = α < α = ∠BAC, the point D lies between B and
C, as shown in the drawing.
Since AD ∼
= A B ∼
= AB by assumption, triangle ABD is isosceles, with baseline
BD. By Euclid I.5, the two base angles of an isosceles triangle are congruent. One of
them is the angle β = ∠ABD. By Proposition 5.36(ii), these base angles are always
acute.
The angle β = ∠A B C ∼
= ∠ADC is the supplement to the second base angle
∼
∠ADB = ∠ABD = β. Hence β and β are congruent to supplementary angles. The
first one is acute, the second one is obtuse, as claimed in (b).
Corollary 11 (SSAA Congruence). Two triangles which have two pairs of congruent
sides, and two pairs of congruent angles across to the latter, are congruent.
Proposition 5.52 (Restricted SSA Congruence, case of a unique solution). As
in the matching Proposition above, we assume that two sides and the angle across one
of these sides of ABC can be matched to congruent pieces of A B C :
(SSA)
AB ∼
= A B , AC ∼
= A C and ∠ACB ∼
= ∠A C B Under the additional assumption that either (i) or (ii) hold,
(i) The given angle γ lies across a side longer or equal to the other given side: c =
AB ≥ AC = b.
(ii) The given angle γ = ∠ACB is right or obtuse.
congruence holds among all triangles matched by (SSA). In all these cases there do
not exist any non congruent triangles with congruent sides b, c and angle γ. Actually
assumption (ii) implies (i).
Proof of Proposition 5.52. Assume that (ii) holds. Indeed, by Proposition 5.36(i), a
triangle can have at most one angle which is right or obtuse. Hence assumption (ii)
implies γ ≥ β. By Euclid I.19, across the greater angle of any triangle lies the longer
side. Hence γ ≥ β implies c ≥ b, which is assumption (i).
We now assume that c = AB ≥ AC = b holds, as stated by (i). By Euclid I.18,
across the longer side of a triangle lies the greater angle. Hence c ≥ b implies γ ≥ β.
A triangle cannot have two angles which are right or obtuse. Hence β and β are both
acute or right. Now congruence follows because case (b) in the matching Proposition is
ruled out.
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Corollary 12 (SSA Congruence for isosceles and right triangles). (i’) Two isosceles triangles with congruent legs and congruent base angles are congruent.
(ii’) Any two right triangles with congruent hypothenuses and one pair of congruent legs
are congruent.
Proof. Assumption (i’) is a special case of (i), and (ii’) a special case of (ii).
Proposition 5.53 (Restricted SSA Congruence Theorem, case with non uniqueness). Again we assume that two sides and the angle across one of these sides of ABC
can be matched to congruent pieces of A B C :
(SSA)
AB ∼
= A B , AC ∼
= A C and
∠ACB ∼
= ∠A C B Complementary to Proposition 5.52, we assume that neither (i) nor (ii) nor β ∼
= R holds.
In other words, we assume that the given angle γ = ∠ACB is acute and lies across the
shorter given side: c = AB < AC = b, and angle β is not a right angle. Under these
additional assumptions (a) There exist two non congruent triangles matched by (SSA).
(b) Nevertheless, equivalent are
(1) The two given triangles ABC and A B C are congruent.
(2) (SSAA) The two triangles can be matched in two sides and the two angles across
both given sides.
(3) For both triangles, the angles β and β across the longer given side b are both acute
or both right or both obtuse.
(4) Both triangles are acute, or both are right, or both triangles are obtuse with the two
obtuse angles at corresponding vertices.
In this case just the two given triangles ABC and A B C —not all other triangles
with those given angle and sides γ, c, b—are congruent.
Proof for part (a). We drop the perpendicular from vertex A onto the opposite side
CB. The foot point F = B, because otherwise β would be a right angle. Transfer
−−→
segment F B on the ray opposite to F B to produce segment F B ∼
= F B. The two
right triangles ABF and AB F are congruent by the Hypothenuse-Leg Theorem
15. Hence AB ∼
= AB . Thus we got two non congruent triangles ABC and AB C
which nevertheless satisfy
(SSA”)
AB ∼
= AB , AC ∼
= AC and ∠ACB ∼
= ∠ACB For these non congruent triangles, the two angles β and β are supplementary. One of
them is acute, the other one is obtuse.
200
Figure 5.68: Establish two solutions for SSA
Proof for part (b). Clearly (1) implies (2) implies (3) implies (4). If (4) holds, the
matching Proposition excludes non congruent solutions.
Indeed, if both triangles are acute, both angles β and β are acute. Congruence
follows by the SSA matching Proposition. If both triangles are right, both angles β
and β are acute or right. In both cases, congruence follows by the SSA matching
Proposition.
If both triangles are obtuse, and obtuse angle occurs at corresponding vertices, then
either both angles β and β are obtuse, or both angles α and α are obtuse. In the second
case both angles both angles β and β are acute. In both cases, congruence follows by
the SSA matching Proposition.
Remark. Earlier on, we have carefully studied triangles with
c = AB = 5 , 5.5 . 6
b = AC = 10
and γ = ∠ACB = 30◦
in Euclidean geometry. In the first case c = 5, one gets a unique right triangle as
solution. In the second case c = 5.5 and c = 6, there are two non-congruent solutions,
one of which is an acute triangle, the other one is an obtuse triangle with β > R.
In the third case c = 6, there are two non-congruent solutions again. But both are
obtuse triangles! One gets one triangle with α > R and β < R, as well as a second one
with α < R and β > R.
5.9
Reflection
Definition 5.14 (Reflection across a Line). Given is a line l. The reflection across
line l maps an arbitrary point P to a point P meeting the following requirements:
If P lies on the symmetry axis l, we put P := P .
If P does not lie on the symmetry axis, the point P is specified by requiring
(i) P and P lie on different sides of l.
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(ii) the lines l and P P are perpendicular, intersecting at F .
(iii) P F ∼
= P F .
Proposition 5.54. Given are two points P, Q on the same side of l and their images
P , Q . We show that
(a) P Q ∼
= P Q
←→
(b) If the lines m := P Q and l intersect at M , then P , Q and M lie on a second line
m .
Question. For this exercise, you can if you like, do drawing and proof, in neutral geometry, on your own.
Figure 5.69: Reflection by a line l
Proof of (a). Let F and G be the foot points of the perpendiculars dropped from P and
Q onto the symmetry axis l. We use SAS-congruence for the two triangles QGF and
Q GF .
Question. Indeed both triangles have a right angle at G—. Explain why.
Answer. ∠QGF is right by definition of a reflection. Since a right angle is congruent to
its supplementary angle, the supplementary angle ∠Q GF is a right angle, too.
—a common side GF and congruent sides GQ ∼
= GQ , by definition of a reflection.
Hence SAS congruence implies
(1)
QGF ∼
= Q GF
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Because of congruence (1), we get
α = ∠QF G ∼
= ∠Q F G ,
QF ∼
= Q F
(2)
(3)
Since ∠GF P ∼
= ∠GF P is a right angle, angle subtraction yields
β = ∠QF P ∼
= ∠Q F P (4)
By construction
PF ∼
= P F
(5)
Using (3)(4) and (5), SAS-congruence implies that
QF P ∼
= Q F P (6)
Hence especially (a) holds.
Proof of (b). We use SAS-congruence for the two triangles QGM and Q GM . Indeed both triangles have a right angle at G: ∠QGM is right by definition of a reflection.
Since a right angle is congruent to its supplementary angle, the supplementary angle
∠Q GM is a right angle, too. Furthermore, they have a common side GM and congruent
sides GQ ∼
= GQ , by definition of a reflection. Hence SAS congruence implies
QGM ∼
= Q GM
For the angle x between the line l of reflection and the given line m, one gets
(8)
x = ∠QM G ∼
= ∠Q M G
Now we argue similarly for the two triangles P F M and P F M . Thus we get
(9)
y = ∠P M F ∼
= ∠P M F = ∠P M G
Now by assumption, the three points P, Q and M lie on a line m.
(10)
x = ∠QM G = ∠P M F = y
From (8)(9),(10) and transitivity of angle congruence, we get
(11)
∠Q M G = ∠P M G
−−→ −−→
Now by Hilbert’s axiom, the angle transfer produces a unique ray. Hence M Q = M P .
Thus the uniqueness of angle transfer implies that the three points M, P and Q lie on
one line.
203
An alternative for the proof of (b) . (6) above implies
(7)
γ = ∠QP F ∼
= ∠Q P F
We apply the extended ASA congruence to triangle P F M and segment F P . The
angles to be transferred to the endpoints of that segment are a right angle at F and γ
at P . Transferring produces as second sides of these angles part of the line l and the
−−→
ray P Q . Hence those rays intersect, say at point M . By ASA-congruence
M F P ∼
= M F P Hence F M ∼
= F M . The points Q and Q and hence M and M lie on the same side of
P P . Hence uniqueness of segment transfer implies M = M as to be shown.
Theorem 5.1. The mapping of reflection preserves incidence and maps segments and
angles to congruent ones. Hence it is an isometry.
5.10
Restriction of SAS congruence
The SAS axiom restricted to triangles of the same orientation is discussed by Hilbert in
an appendix to the Foundations of Geometry.
Note that the triangles ABC and A B C have the same orientation if either
−→
both point C lies in the left half-plane of ray AB and point C lies in the left half-plane
−−→
−→
of ray A B ; or both point C lies in the right half-plane of ray AB and point C lies in the
−−→
right half-plane of ray A B . Hilbert postulates SAS congruence restricted to triangles
with the same orientation. and two additional axioms:
III.5* Assume that the two triangles ABC and A B C have the same orientation
and the congruences
AB ∼
= A C , ∠BAC ∼
= ∠B A C = A B , AC ∼
hold, then the congruence ∠ABC ∼
= ∠A B C is also satisfied.
III.6 If the angles ∠(h , k ) and ∠(h , k ) are both congruent to angle ∠(h, k), then the
angles ∠(h , k ) and ∠(h , k ) are congruent to each other.
III.7 If the rays c and d have the same vertex as the angle ∠(h, k) and lie in the interior
of that angle, then the angles ∠(h, k) and ∠(c, d) are not congruent.
Hilbert then announces the following result without a proof:
Proposition 5.55 (The importance of Euclid I.5). We assume only the restricted
SAS axiom (III.5*), axioms (III.6) and (III.7), and finally we assume Euclid (I.5) to be
valid: the base angles of any isosceles triangle are congruent. These assumptions imply
the unrestricted SAS axiom (III.5),—for any triangles independent of their orientation.
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Problem 5.11 (A hard problem, only for fans). Provide some drawings and prove
this proposition.
Definition 5.15 (Skew-Hilbert plane). A skew-Hilbert plane is any model for twodimensional geometry where Hilbert’s axioms of incidence (I.1)(I.2)(I.3a)(I.3b), order
(II.1) through (II.4), and congruence (III.1) through (III.4) , the restricted SAS-axiom
(III.5*), and the axioms (III.6) and (III.7) hold.
Lemma 5.7 (Transfer of a triangle with the same orientation). Any given triangle
ABC can be transferred into either the left or the right half-plane of any given ray r,
to obtain a congruent triangle with one vertex at the vertex of this ray, one side lying
on the given ray, and having the same orientation as the original triangle.
Lemma 5.8 (Preparations in a skew-Hilbert plane). The following Propositions are
valid in any skew-Hilbert plane
• The angle congruence is an equivalence relation. Too, Proposition 5.24 holds:
angle comparison holds for congruence classes.
• Proposition 5.10 about SAS congruence holds for two triangles with the same orientation.
• As stated in Lemma 5.7, the transfer of a triangle is still possible under the restriction that the orientation is preserved.
• Proposition 5.18 about angle-addition and subtraction (Theorem 15 in Hilbert)
remains valid.
• The Exterior Angle Theorem (Euclid (I.16). Theorem 22 in Hilbert, see Proposition 5.34) is valid.
• The SAA-Congruence Theorem, Proposition 5.45 (Theorem 25 in Hilbert) is valid
for triangles of the same orientation.
• The midpoint of a segment can be obtained by the method used in Theorem 26 of
Hilbert (see Proposition 5.49). Any segment has a midpoint.
Lemma 5.9. The following Propositions are valid in a skew-Hilbert plane where additionally Euclid (I.5) is assumed.
• From the Theorem 5.19 about the symmetric kite, part (i) holds.
27
• Euclid(I.18) (Theorem 23 of Hilbert, see Proposition 5.39 above) is valid: in any
triangle, across the longer side lies the greater angle.
27
Part (ii) cannot be recovered at this point but only later.
205
• Euclid (I.19) (Proposition 5.40 above) is valid: in any triangle, across the greater
angle lies the longer side.
• We get Euclid (I.6), the converse of Euclid (I.5).
Figure 5.70: To get a rhombus without use of reflection.
Lemma 5.10. The diagonals of the rhombus bisect each other, they bisect the angles
at the vertices of the rhombus, and they are perpendicularly to each other. For any
isosceles triangle ABC, we can construct a rhombus ABCD which inherits three
of its vertices from the triangle.
Proof of the Lemma. The midpoint M of side AC can be obtained with Hilbert’s construction. From SAS axiom (III.5*) we derive the triangle congruences
AM B ∼
= CM B and
CM B ∼
= AM B Hence we have obtained a rhombus ABCB . Too, get four congruent angles at vertices
A and C and see that the diagonal AC bisects the angles at these two vertices. Beginning
with the isosceles triangle BAB we repeat a similar argument. Hence the diagonals
of a rhombus bisect each other, and they bisect the angles at the vertices of the rhombus.
206
It remains to check that the diagonals are perpendicular to each other. To this
end, we transfer the hypothenuse BC of triangle BM C on the segment BA and
obtain the congruent triangle BM A ∼
= M BC with the same orientation. One needs
only axioms (III.1),(III.4) and the restricted SAS axiom (III.5*). From the first item
(i) of the Theorem 5.19 about the symmetric kite, we conclude ∠BM A ∼
= ∠BM A.
From the triangle congruence we conclude we conclude ∠BM C ∼
= ∠BM A. Hence
∠BM C ∼
= ∠BM A are congruent supplementary, and hence right angles.
Lemma 5.11. All points P of the perpendicular bisector of two points B and C have
congruent distances BP ∼
= CP from these two points.
Proof. We may suppose that point P does not lie on the line BC. We take the smaller
one of the two angles ∠CBP and ∠BCP and transfer it to the vertex and onto the side
of the other one. Suppose to be definite that ∠BCP ≤ ∠CBP . We transfer ∠BCP onto
−−→
the ray BC. By the Crossbar Theorem, the newly produced ray intersects the segment
BP , say at point Q. The triangle CQB has congruent base angles, and hence from
Euclid (I.6) we conclude BQ ∼
= CQ.
It remains to show that Q = P . Let M be the midpoint of segment BC. From the
Proposition about the rhombus (Lemma 5.10) we know that line QM is the perpendicular bisector of segment BC. Both P and Q are the intersection point of the two
different lines QM and BP , hence P = Q. Finally we have confirmed BP ∼
= CP .
Lemma 5.12. In a skew-Hilbert plane Euclid (I.5) has been assumed. Theorem 5.19
about the symmetric kite, parts (i) and (ii) both hold. Indeed, for a kite with congruent
sides XZ1 ∼
= XZ2 and Y Z1 ∼
= Y Z2 , and Z1 and Z2 on different sides of XY , the
triangles XZ1 Y ∼
= XZ2 Y are congruent.
End of the proof of the Theorem 5.19 about the symmetric kite. Let Z1 and Z2 be two
points on different sides of line XY , and assume that XZ1 ∼
= XZ2 and Y Z1 ∼
= Y Z2 .
As explained above, from the isosceles triangle Z1 XZ2 , we construct a rhombus
Z1 XZ2 X . Its diagonals XX and Z1 Z2 intersect perpendicularly in the midpoint M of
segment Z1 Z2 .
We begin a similar procedure with the isosceles triangle Z1 Y Z2 and obtain a
second rhombus Z1 Y Z2 Y . Its diagonals Y Y and Z1 Z2 intersect perpendicularly in the
midpoint M of segment Z1 Z2 .
Hence all four points X, X , Y, Y lie on the perpendicular to segment Z1 Z2 , erected
at its midpoint M ,—which we recognize as the perpendicular bisector of Z1 Z2 . Since
we have shown that the diagonals XX and Y Y bisect the angles at X and Y , we get
the remaining angle congruences for triangles XZ1 Y ∼
= XZ2 Y .
Lemma 5.13. Assume that the triangles AP B and AP C lie on different sides of
AP , have congruent sides AB ∼
= AC and congruent angles ∠P AB ∼
= ∠CAP between
these sides and their common side. Then the two triangles are congruent.
207
Proof. From the Proposition about the rhombus (Lemma 5.10) we know that line AP
is the perpendicular bisector of segment BC. From Lemma 5.11, we know that BP ∼
=
CP . From the Theorem 5.19 about the symmetric kite we conclude that the triangles
AP B ∼
= AP C are congruent.
End of the proof of the Proposition about importance of Euclid (I.5). Combine Lemma 5.13
with the Lemma 5.7 about the transfer of a triangle. Assume the two triangles ABC
and A B C have different orientations, have corresponding congruent sides AB ∼
= A B and AC ∼
= A C , and congruent angles ∠BAC ∼
= ∠B A C between them.
−−→
We transfer the angle ∠BAC onto the ray A C , at vertex A , to the same side of
A C as B . On the newly produced ray, we transfer segment AB, starting at vertex A .
Thus we get point B0 , such that A B0 ∼
= AB. Because of Lemma ?? we get
ABC ∼
= A B0 C By Lemma 5.13 we conclude that
A B C ∼
= A B0 C Hence ABC ∼
= A B C as to be shown.
208