Download 3 Solution of Homework

Math 3181
Name:
Dr. Franz Rothe
February 25, 2014
All3181\3181_spr14h3.tex
Homework has to be turned in this handout.
The homework can be done in groups up to three
due March 11/12
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Solution of Homework
10 Problem 3.1. Give a detailed definition what is meant by congruence of two
generic triangles. Moreover, explain with exact notation what the congruence 4ABC ∼
=
0 0 0
4A B C means,—illustrated with a colored figure.
Answer.
Definition 1 (Congruence of triangles). Two triangles are called congruent iff all
three pairs of corresponding sides, and all three pairs of corresponding angles are congruent. Thus the congruence of triangles 4ABC ∼
= 4A0 B 0 C 0 means the six congruences
(3.1)
AB ∼
= A0 B 0 , BC ∼
= B 0 C 0 , CA ∼
= C 0 A0 and
∠ABC ∼
= ∠A0 B 0 C 0 , ∠BCA ∼
= ∠B 0 C 0 A0 , ∠CAB ∼
= ∠C 0 A0 B 0
Note that one has to put the vertices, sides and angles of the two triangles into a definite
order.
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Figure 1: How to simply get ASA congruence
Proposition 1 (ASA Congruence). [Theorem 13 in Hilbert] Two triangles with a
pair of congruent sides, and pairwise congruent adjacent angles are congruent.
10 Problem 3.2. State the ASA congruence, with the notation from the figure
above. Prove the theorem.
Answer (Independent proof of ASA congruence). Given are the triangles 4ABC and
4A0 B 0 C 0 , with congruent sides AB ∼
= A0 B 0 and two pairs of congruent adjacent angles
0
0
0
∠BAC ∼
= ∠B A C and ∠ABC ∼
= ∠A0 B 0 C 0 . The segment AC is transferred onto the
−−
→
ray A0 C 0 . On this ray, one gets a point X such that
(3.2)
AC ∼
= A0 X
Now axiom (III.5) is applied to the triangles 4ABC and 4A0 B 0 X. In the figure on
page 2, the three pairs of matching pieces used to prove the congruence are stressed in
matching colors. From axiom (III.5) one concludes
(3.3)
∠ABC ∼
= ∠A0 B 0 X
We now have obtained both ∠ABC ∼
= ∠A0 B 0 X, and ∠ABC ∼
= ∠A0 B 0 C 0 as assumed,
too. Hence uniqueness of angle transfer, postulated by axiom (III.4b), implies the rays
to be equal:
−−0→ −−
→
B X = B0C 0
−−→ −−→
Thus the point X lies both on this ray, and by construction on the ray A0 X = A0 C 0 , too.
By axiom (I.2), these two rays have a unique intersection point, since they do not lie on
the same line, and hence X = C 0 . By substitution of equals, equations (2.2) and (2.3)
imply
(3.4)
AC ∼
= A0 C 0 and ∠ABC ∼
= ∠A0 B 0 C 0
Finally we show BC ∼
= B 0 C 0 by a similar argument with vertices A and B exchanged.
Hence the two triangles are congruent, as to be shown.
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Proposition 2 (SAS Congruence). [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.
10 Problem 3.3. Complete the following short proof of the SAS-congruence
Proposition 2.
Short proof of SAS-congruence Proposition 2. Given are two triangles !! 4ABC and 4A0 B 0 C 0
. We assume that the angles at A and A0 as well as two pairs of adjacent sides are
matched:
AB ∼
= A0 B 0 , AC ∼
= A0 C 0 , ∠BAC ∼
= ∠B 0 A0 C 0
By axiom !! (III.5)
we conclude
∠ABC ∼
= ∠A0 B 0 C 0
Now the !! ASA -congruence stated in !! Proposition 1
congruence 4ABC ∼
= 4A0 B 0 C 0 to be shown.
above yields the triangle
Question. We see from Proposition 2 that Hilbert’s SAS-axiom (III.5) implies the full
SAS congruence theorem. 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.
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Figure 2: Transfer of a triangle.
Proposition 3 (Transfer of a triangle). Any given triangle 4ABC can be transferred
into any given half-plane H0 of any given ray r, such that one obtains a congruent
−−→
triangle 4A0 B 0 C 0 lying in the prescribed half-plane, and the given ray is r = A0 B 0 ,
emanates from vertex A0 and lies on the side A0 B 0 .
10 Problem 3.4. Use the notation from the figure above. Prove the Proposition 3,
starting from the axioms and ASA-congruence. State clearly which axioms you use, and
where one has to use the ASA-congruence.
Answer. Proof. Using axiom (III.1), we transfer the segment AB onto the ray r to
obtain the congruent segment AB ∼
= A0 B 0 . Next we axiom (III.4), and transfer the angle
∠BAC into the given half-plane H0 to obtain the congruent angle ∠BAC ∼
= ∠(r, k).
Thus the newly produce angle has the given ray r as one side. Onto the other side k
of the newly produced angle, we transfer the segment AC and obtain the congruent
−−→
−−→
segment AC ∼
= A0 C 0 . Since by construction r = A0 B 0 and k = A0 C 0 , we see that
∠BAC ∼
= ∠B 0 A0 C 0 .
From axiom (III.5)—which we now use even literally!—we conclude the congruence
∠ABC ∼
= ∠A0 B 0 C 0 . We are now in the position to use the ASA-congruence, stated
in Proposition 1 for the triangles 4ABC and 4A0 B 0 C 0 . Hence we conclude that the
triangle congruence 4ABC ∼
= 4A0 B 0 C 0 , as to be shown.
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Proposition 4 (Extended ASA-Congruence). Given is a triangle, a segment congruent to one of its sides, and a half-plane H0 bounded by this segment . 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.
Figure 3: Extended ASA congruence
10 Problem 3.5. Complete the following short proof of the extended ASA-congruence
Proposition 4.
Short proof of the extended ASA-congruence Proposition 4. Given are the triangle !! 4ABC
−−→
and the segment A0 B 0 ∼
= AB. We define the ray r = A0 B 0 . By Proposition 3 about
!! the transfer of a triangle , we transfer the triangle 4ABC into the given half-plane
H0 . We obtain a congruent triangle 4A0 B 0 C 0 with vertex C 0 lying in the prescribed
half-plane.
Question. 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.
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Proposition 5 (The diagonals of the rhombus bisect each other perpendicularly). Given are four different points A, B, C, D lying in one plane such that the
segments AB ∼
= BC ∼
= CD ∼
= DA are congruent. Then the segments AC and BD
bisect each other perpendicularly at their common midpoint.
Figure 4: The diagonals of the rhombus bisect each other.
Question. Why do points A and C lie on different sides of line BD, points B and D lie
on different sides of the other diagonal AC.
Answer. 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.
Question. Give a reason why the left- and right triangles 4ABD ∼
= 4CBD are congruent (see third figure).
Answer. 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 4ABD ∼
4CBD
are
con=
gruent (see third figure).
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Furthermore, the latter two triangles are both isosceles. Hence 4ABD ∼
= 4CDB
∼
holds, too, and ∠ABD = ∠CDB (see first figure in the second row). Similarly, we
prove ∠CAB ∼
= ∠ACD. The two segments AC and BD intersect. Let point M be
their intersection point.
Question. How does one get the congruence 4M AB ∼
= 4M CD (see second figure in
the second row). Why do the diagonals AC and BD bisect each other (see last figure
in the second row).
Answer. By ASA congruence, we get 4M AB ∼
= 4M CD (see second figure in the
second row). Hence the diagonals AC and BD bisect each other (see last figure in the
second row).
Question. Justify the triangle congruence 4M AB ∼
= 4M CB.
Answer. Since the isosceles triangle 4ABC has congruent base angles, SAS congruence
yields 4M AB ∼
= 4M CB.
Question. Why are the diagonals AC and BD are perpendicular to each other.
Answer. Since 4M AB ∼
= 4M CB, 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.
10 Problem 3.6. Give purely geometric definitions for supplementary angles
and for vertical angles. (Do not use any measurements!)
Answer.
Definition 2 (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 3 (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.
10 Problem 3.7. Give purely geometric definitions for right angle, acute angle
and for obtuse angle. (Use only comparison of angles, not any measurements!)
Answer.
Definition 4 (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.
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Figure 5: Four right triangles yield a kite. When is it even a rhombus?
Proposition 6 (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.
10 Problem 3.8. Complete the following proof of the hypothenuse leg theorem.
First proof of the hypothenuse-leg theorem 6 . Given are two right triangles 4ABC and
4A0 B 0 C 0 . As usual, we put the right angles at vertices C and C 0 . We assume congruence
of the hypothenuses AB ∼
= A0 B 0 and of one pair of legs AC ∼
= A0 C 0 .
We build a kite out of two copies of triangle 4ABC and two copies of triangle
−→
4A0 B 0 C 0 . To this end, one !! transfers angle ∠C 0 A0 B 0 onto the ray AC into the opposite half plane and gets by transfer of triangle (see Proposition 3) the congruence
4C 0 A0 B 0 ∼
= 4CAB 00 . Especially
(3.5)
B0C 0 ∼
= B 00 C
Since all right angles are congruent (see Proclus’s Proposition), the two supplementary
right angles at vertex C imply that points !! B, C and B 00 lie on a line.
−→
Next we transfer segment CA onto the ray opposite to CA and get point D such that
askCA ∼
= CD. We produce the two triangles to the right, as shown in the second row
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of the figure on page 8. Using the right angle, !! SAS congruences imply 4ABC ∼
=
00
00
∼
4DBC and !! 4AB C = 4DB C .
Since AB ∼
= A0 B 0 ∼
= AB 00 and DB ∼
= AB ∼
= AB 00 ∼
= DB 00 , one has constructed a
symmetric kite with the four points
X := A, Y := D, Z1 := B, Z2 := B 00
We can apply Hilbert’s kite-theorem 7 and conclude
∠DAB = ∠XY Z1 ∼
= ∠XY Z2 = ∠DAB 00
The congruence of the same angles is ∠CAB = ∠CAB 00 . Hence SAS congruence implies
now !! 4CAB ∼
= 4CAB 00 and hence
(3.6)
BC ∼
= B 00 C
Together, the formulas (2.5) and (2.6) imply B 0 C 0 ∼
= BC. Since all right angles are
congruent, a final SAS congruence using the right angles implies congruence 4ABC ∼
=
0 0 0
4A B C of the originally given triangles, as to be shown.
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Proposition 7 (The Kite-Theorem). [Theorem 17 in Hilbert] Let Z1 and Z2 be two
points on different sides of line XY , and assume that XZ1 ∼
= XZ2 and Y Z1 ∼
= Y Z2 .
∼
Then the two triangles 4XY Z1 = 4XY Z2 are congruent.
10 Problem 3.9. Complete the proof of the kite-theorem and provide drawings
for all three cases.
Proof. The congruence of the !! base angles of the !! isosceles triangle 4XZ1 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 the angle ∠XZ1 Y , angle !! subtraction
−−−→
in case ray Z1 Z2 lies !! outside angle ∠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.
One now applies SAS congruence to 4XZ1 Y and !! 4XZ2 Y . Indeed the angles at
Z1 and Z2 and the adjacent sides are pairwise congruent. Hence the assertion 4XY Z1 ∼
=
4XY Z2 follows.
Answer. Here are drawings for the three cases.
Figure 6: The symmetric kite
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