Download 8 Standard Euclidean Triangle Geometry

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8.1
Standard Euclidean Triangle Geometry
The circum-center
Figure 8.1: In hyperbolic geometry, the perpendicular bisectors can be parallel as shown—
but this figure is impossible in Euclidean geometry.
Theorem 8.1 (The Circum-center). In Euclidean geometry, the bisectors of all three
sides intersect in one point— which is the center of the circum-circle.
Reason. One needs to check that two perpendicular bisectors really intersect. Then the
existence of a circum-circle follows by Proposition 9.2.
We assume towards a contradiction that all three perpendicular bisectors are parallel.
By plane separation, they partition the plane into four strips. Two of the vertices, say
A and C, lie in half planes which are bounded by just one bisector —the most left and
most right in figure 8.1. The third vertex B lies in either
(a) one of the two strips in the middle—or
(b) on the middle bisector.
Assume case (b) occurs and bisector ob of side AC goes through vertex B. By Pasch’s
axiom each perpendicular bisector of a triangle side intersects two sides of the triangle.
For the triangle Mb CB, the bisector oa intersects side BC, but not side B, Mb . Hence
bisector oa intersects the segment Mb C, say at point Pa . The right triangle Ma Pa C
has an acute angle at vertex Pa .
Hence the two bisectors ob and oa intersect the side AC at a right, and at an acute
angle, respectively. By Euclid I.28—or rather literally the fifth postulate!— the two lines
ob and oa do intersect. Now the existence of a circum-circle follows by Proposition 9.2.
Assume case (a) occurs. By Pasch’s axiom each perpendicular bisector of a triangle
side intersects two sides of the triangle. In the situation set up in the beginning, side
AC is intersected by three bisectors. Let the perpendicular bisector of side AB be
oc = Mc Pc , the perpendicular bisector of side BC be oa = Ma Pa , were points Pc and Pa
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lie on the segment AC. The angle χ = ∠APc Mc is acute, because it is an angle of right
triangle AMc Pc . The angle ϕ = ∠APa Ma is obtuse, because it is an exterior angle of
right triangle Pa Ma C.
The perpendicular bisectors oc and oa intersect line AC at z-angles χ and ϕ—the
first one is acute, and the other one obtuse. Hence, once more, we conclude that these
two bisectors do intersect—leading to a contradiction. Once more, existence of a circumcircle follows by Proposition 9.2.
Note of caution. In hyperbolic geometry, both cases (a) and (b) explained above cannot
be ruled out, but are perfectly possible.
8.2
Double and half size triangles
Figure 8.2: Constructing a triangle of double size.
Proposition 8.1 (Double ASA). Given is a triangle A B C and a segment AB =
2A B , of double length than A B . Then there exists a triangle ABC, equiangular with
A B C . All three sides have length double as the corresponding sides of the original
triangle.
Reason. Let M be the midpoint of segment AB, and hence A B ∼
= AM ∼
= M B. We
stay now in one half plane of line AB, and transfer the angle β = ∠A B C onto both
−→
−→
−−→
rays BA and M A. Similarly, we transfer the angle α = ∠B A C onto the two rays AB
−−→
and M B. By the extended ASA congruence one gets two intersection points Mb and
Ma . The two new triangles and the one to begin with are all congruent.
(8.1)
A B C ∼
= AM Mb ∼
= M BMa
Since we deal with Euclidean geometry, the angle sum of a triangle is 2R (Euclid I.32).
Hence angle addition at vertex M implies ∠Ma M Mb ∼
= γ. Now SAS congruence implies
(8.2)
A B C ∼
= Ma Mb M
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Hence congruent z-angles (Euclid I.27) yield that the lines AB and Ma Mb are parallel.
−−→
The extensions of the already produced ray AMb intersect both parallel segments AB
and Ma Mb at congruent angles α. (Here we need to use Euclid 1.28 and hence the
Euclidean parallel property!)
−−−→
Similarly, the extensions of the already produced ray BMa intersect both parallel
−−→
segments AB and Ma Mb at angles β. By the extended ASA congruence, the rays AMb
−−−→
and BMa do intersect, say at point C, and
(8.3)
A B C ∼
= Mb Ma C
Altogether, the congruences (1.11),(1.11) and (9.7) have produced four new congruent
triangles. Furthermore Ma , Mb and M are the midpoints of the sides of the larger new
triangle ABC.
Proposition 8.2 (Half ASA). Given is a triangle ABC, and a segment A B of half
length as AB. Then there exists a triangle A B C , equiangular with ABC, and all
three sides of it have half length as the sides of the original triangle.
Reason. The half segment A B is smaller than the original segment AB. By my earlier
remarks about extended ASA and Pasch’s axiom in the section on Neutral Geometry ??,
A B < AB implies the existence of a triangle A B C such that
∠ABC ∼
= ∠A B C and ∠BAC ∼
= ∠B A C By Euclid I.32 both triangles ABC and A B C have the same angle sum 2R. Hence
two pairs of congruent angles are enough to show that they are equiangular.
Now we are back to the situation of Proposition 8.1, which is applied to the smaller
triangle A B C and the (larger) segment AB.
Based on the segment AB, we construct a triangle ABD, which is equiangular with
triangle A B C and hence with the original triangle ABC. Because of ABC ∼
=
ABD, uniqueness of angle transfer yields D = C. As expected, we got back the
original triangle.
By Proposition 8.1, the sides of triangle ABD = ABC are double the sides
of A B C . Of course, this means that the sides of A B C are half the sides of
ABC.
Proposition 8.3 (The midpoint triangle). The midpoints of the sides of a triangle
and the segments between them produce four smaller congruent triangles. They are
equiangular to the original triangle, and have sides half as long as the corresponding
sides of the original triangle.
”Halving then doubling gives back the original triangle”. We draw the parallel to side
BC through point Mc . By Pasch’s axiom, the parallel intersects segment AC, say in
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Figure 8.3: Construction and properties of the midpoint triangle.
point Pb . Furthermore, the ”halved” triangle AMc Pb is equiangular to the original
triangle ABC.
By Proposition 8.2, all three sides of AMc Pb are half of the corresponding sides of
ABC. Hence AC = 2APb and Pb = Mb is the midpoint of side AC. This means that
the parallel to BC through point Mc does intersect triangle side AC in its midpoint.
Similarly, one sees that the other two sides of the midpoint-triangle Ma Mb Mc are
pairwise parallel to the sides of the original triangle. By Euclid I.28, a transversal
crossing two parallel lines produces congruent z-angles. (This is a strong statement,
valid only in Euclidean geometry!). Hence the midpoint triangle partitions the original
triangle into four smaller, half-size triangles, which are all equiangular to the original
triangle.
Proposition 8.4 (Double SAS). Assume that the two triangles ABC and A B C have congruent angles ∠BAC ∼
= B A C , and the corresponding adjacent sides satisfy
AB ∼
= 2A B and AC ∼
= 2A C . Then they are equiangular, and BC ∼
= 2B C .
Reason. This follows easily from Double ASA:—as the reader should check—:
By Double ASA, there exists a triangle ABC∗ , which is equiangular to the given
triangle and all three sides are of double length than the original triangle A B C .
Hence AC∗ ∼
= 2A C . On the other hand, AC ∼
= 2A C was assumed. Hence axiom
−→
III.2 yields AC∗ ∼
= AC. Reproducing this segment on the ray AC yields a unique point
C = C∗ . 41
8.3
The centroid
Theorem 8.2 (The Centroid). The three medians of a triangle intersect in one point,
called the centroid. The centroid divides the medians in the ratio 2 : 1.
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If you still need a bottle of wine from Hilbert:
Applying the SAS axiom III.5 to the two triangles ABC and ABC∗ , we conclude ∠ABC ∼
=
−−→ −−→
∠ABC∗ . Hence the uniqueness of angle transfer stated in axiom III.3 implies r = BC = BC∗ . Hence
−→
C = C∗ is the unique intersection point of the two rays r and AC.
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Figure 8.4: The three medians intersect in the centroid.
Reason, given for Euclidean geometry. Apply the Double SAS Proposition 8.4 to the
two triangles, AMb Mc and ABC. We see that they are equiangular, and that the
side BC ∼
= 2Mb Mc . Hence the segment Mb Mc is parallel to the side BC.
At first, we define point S as intersection of the two medians BMb and CMc . The
triangles SMb Mc and SBC are equiangular. We can apply the Double ASA Proposition 8.1 to the triangle SMb Mc and the (longer) segment BC. Hence we conclude
that SB = 2SMb and SC = 2SMc .
By the fact of dividing the median BMb as 2 : 1, the point S on median BMb is
uniquely defined. Now define point T as the intersection point of medians BMb and
AMa . The same argument as above shows that point T divides the median BMb as
2 : 1, too. Hence S = T is the intersection of all three medians.
8.4
The orthocenter
Theorem 8.3 (The Orthocenter). In Euclidean geometry, the three altitudes of a
triangle intersect in one point—which is called the orthocenter.
Concise proof. We can double one side of the given triangle. By Double ASA, Proposition 8.1, one constructs an equiangular triangle A0 B 0 C 0 with all three sides doubled. The original triangle ABC is congruent to the midpoint triangle Am Bm Cm of
A0 B 0 C 0 .
ABC ∼
= Am Bm Cm
The altitudes of the triangle Am Bm Cm are the perpendicular bisectors of the sides
of the larger triangle A0 B 0 C 0 . Hence they intersect in one point. This point is the
circum-center of A0 B 0 C 0 , as well as the orthocenter of Am Bm Cm .
Proposition 8.5. In Euclidean geometry, every triangle ABC is the midpoint triangle
of a larger triangle A0 B 0 C 0 .
Proof I: Use extended ASA congruence three times. Use extended ASA congruence three
times, and get three new triangles:
ABC ∼
= CB 0 A
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Figure 8.5: Every triangle ABC is the midpoint triangle of a larger triangle A0 B 0 C 0 .
lying on opposite sides of AC. Similarly,
CB 0 A ∼
= ABC
are lying on opposite sides of AC. Finally
A0 CB ∼
= ABC
are lying on opposite sides of BC. Because the angle sum is α + β + γ = 2R, point C
lies on segment A0 B0 . Furthermore B0 C ∼
= AB ∼
= CA0 . Hence C is the midpoint of
segment A0 B0 . Similarly, we see that A is the midpoint of segment B0 C0 and B is the
midpoint of segment C0 A0 . hence the given triangle ABC is the midpoint-triangle of
the larger triangle A0 B 0 C 0 .
Proof II: Erect the perpendiculars on the altitudes. One constructs the doubled triangle
A0 B 0 C 0 by erecting the perpendiculars on its altitudes at all three vertices of triangle
ABC. Let pa , pb , pb be these three perpendiculars. Using angle sums, one can check
that
−→
−→
∠(pa , AC) ∼
= γ and ∠(pc , CA) ∼
=α
By the extended ASA congruence, lines pa and pc intersect at a point B 0 , and
ABC ∼
= CB 0 A
These two triangles lie on opposite sides of AC. Similarly, one get a new triangle
CB 0 A ∼
= ABC
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These two triangles lie on opposite sides of AC. Similarly, one get a new triangle
A0 CB ∼
= ABC
on opposite side of BC and
BAC 0 ∼
= ABC
on opposite side of AB. In this way, the intersection points of the three lines pa , pb , pb
yield a larger triangle A0 B 0 C 0 .
Question. What happens if one does the same two constructions in hyperbolic geometry?
Answer. Can one construct the three new congruent triangles outside of the given triangle by the extended ASA theorem. One ends up a figure with four congruent triangles,
but its outer boundary is a convex hexagon, not a triangle.
Alternatively, one can use the double perpendiculars, erected at the vertices on the
altitudes of the given triangle. These three lines need not even intersect! In case they
do intersect, the smaller triangle is indeed the midpoint-triangle of the larger triangle.
But the larger triangle A0 B0 C0 has angles smaller than the angles of the given
triangle ABC.
Constructive proof for the orthocenter—using midpoints. By Proposition 8.5, the given
triangle ABC is the midpoint-triangle of a larger triangle A0 B0 C0 . By Proposition 9.1, the altitudes of the triangle ABC are the perpendicular bisectors of the sides
of the larger triangle A0 B 0 C 0 .
By Proposition 8.1, the perpendicular bisectors intersect in one point. This point is
the circum-center of the larger triangle A0 B 0 C 0 , as well as the the orthocenter of the
original triangle ABC.
Corollary 47. The circum-center of a triangle is the orthocenter of the midpoint triangle.
8.5
The in-circle and the three ex-circles
Theorem 8.4 (The in-circle and three ex-circles). In Euclidean geometry, a triangle has an in-circle and three ex-circles. All four circles have the three sides of the
triangle, or their extensions, as tangents. The in-circle touches the three sides from
inside the triangle. The ex-circles touch one side from outside, and the extensions of
the two other sides.
Reason. The existence of the in-circle holds already in neutral geometry. To get the
ex-circle touching side AB form outside, consider the two exterior bisectors eA and eB
at vertices A and B. They both form acute angles with segment AB, and both lie on
the side opposite to vertex C. By Euclid’s fifth postulate these exterior bisectors do
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intersect, say in point Ic . 42 Now one argues similarly as in the case of the in-circle.
Point Ic has congruent distances to all three sides of the triangles, and hence lies on the
interior bisector of angle ∠BCA, too. Point Ic is the center of the ex-circle touching
side AB from outside, as well as the extensions of the two other sides.
Remark. The bisectors of the exterior angles of the triangle are the perpendiculars to
the angular bisectors, erected at the vertices.
8.6
The road to the orthocenter via the orthic triangle
Definition 8.1 (The orthic triangle). Let Fa , Fb and Fc denote the foot points of
the three altitudes. The triangle Fa Fb Fc is called the orthic triangle.
Figure 8.6: The altitudes are the angular bisectors of the orthic triangle.
Theorem 8.5. For an acute triangle, the altitudes are the inner angular bisectors of
the orthic triangle.
For an obtuse triangle, the only the altitude dropped from the obtuse angle is an inner
angular bisector. The other two altitudes are exterior angular bisectors.
Corollary 48. The three altitudes of a triangle intersect in one point.
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The unique parallel to eA through point B is different to eB .
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The case of an acute triangle. We shall repeatedly use the congruence of circumference
angle, stated by Euclid III.21. Both angle ∠CAFa ∠CBFb ∼
= R − γ because of the angle
sum and the right angles. By the converse Thales theorem, the four points C, Fb , Fc and
B lie on a semicircle with diameter BC. Hence the congruence of circumference angles
(Euclid III.21) implies that ∠CBFb ∼
= ∠CFc Fb .
Similarly, the four points C, Fa , Fc and A lie on a semicircle with diameter AC, and
the congruence of circumference angles implies ∠CAFa ∼
= ∠CFc Fa .
Hence ∠CFc Fa ∼
= ∠CFc Fb , and the altitude CFc bisects the angle ∠Fa Fc Fb of the
orthic triangle. Similarly, we can show that all three altitudes are angular bisectors of
the orthic triangle.
Figure 8.7: The orthic triangle for an obtuse triangle.
The Corollary: Constructive proof for the orthocenter—using bisectors. Since we did not
need the intersection point of any two altitudes, but we know from Proposition 9.3 that
the three angular bisectors intersect in one point, one gets an additional proof that the
three altitude intersect in one point.
Proposition 8.6. In Euclidean geometry, every acute triangle ABC is the orthic
triangle of two non congruent larger triangle, one acute A1 B 1 C 1 , and one obtuse
A2 B 2 C 2 .
Proof. The first good exercise. Do it!
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8.7
The Euler line
Theorem 8.6 (The Euler line). The orthocenter, the centroid, and the circum-center
of a triangle lie on one line, called the Euler line. The centroid trisects the segment
joining the orthocenter and the circum center.
Remark. In the exceptional case of an equilateral triangle, all three centers are equal,
but the Euler line is not defined.
Definition 8.2. A central dilation with center C and ratio k = 0 is a mapping that
maps point A to a point A such that the three points A, A and C lie on a line and
A C ∼
= CA. If k > 0, the center C lies outside the segment AA . If k < 0, the center C
lies inside the segment AA .
Proposition 8.7. A central dilation maps a triangle XY Z to an equiangular triangle X Y Z such that X Y ∼
= kXY , Y Z ∼
= kY Z and Z X ∼
= kZX and these
corresponding segments are parallel.
Partial reason. We shall need only the special case k = −2, which can be covered with
congruence theorems and Double congruence theorems from above.
Proof of Euler’s Theorem. Given is a triangle ABC. We use a central dilation with
the centroid S as center and ratio k = −2.
The Centroid Theorem 8.2—together with the definition of a central dilation—
implies that the midpoints Ma , Mb and Mc are mapped to the vertices A, B and C.
Because of conservation of angles, the altitudes of the midpoint triangle are mapped to
the altitudes of the original triangle. Furthermore, the orthocenter H2 of the midpoint
triangle is mapped to the orthocenter H of triangle ABC.
By Proposition 9.1, the orthocenter H2 of the midpoint triangle is the circum-center
O of triangle ABC.
Hence the dilation maps O to H and, of cause, center S to itself. Directly from the
definition of a central dilation, we see that the three centers O, S and H lie on one line
and 2 OS ∼
= SH, with the centroid S between the circum-center and the orthocenter.
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Figure 8.8: The Euler line comes from a dilation with center S.
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