8 Standard Euclidean Triangle Geometry

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1 8 Standard Euclidean Triangle Geometry 8.1 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 o b 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 M b CB, the bisector o a intersects side BC, but not side B,M b. Hence bisector o a intersects the segment M b C,sayatpointP a. The right triangle M a P a C has an acute angle at vertex P a. Hence the two bisectors o b and o a 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 o b and o a 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 o c = M c P c, the perpendicular bisector of side BC be o a = M a P a,werepointsp c and P a 502

2 lie on the segment AC. The angle χ = AP c M c is acute, because it is an angle of right triangle AM c P c. The angle ϕ = AP a M a is obtuse, because it is an exterior angle of right triangle P a M a C. The perpendicular bisectors o c and o a 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 = MB. We stay now in one half plane of line AB, and transfer the angle β = A B C onto both rays BA and MA. Similarly, we transfer the angle α = B A C onto the two rays AB and MB. By the extended ASA congruence one gets two intersection points M b and M a. The two new triangles and the one to begin with are all congruent. (8.1) A B C = AMMb = MBMa Since we deal with Euclidean geometry, the angle sum of a triangle is 2R (Euclid I.32). Hence angle addition at vertex M implies M a MM b = γ. Now SAS congruence implies (8.2) A B C = Ma M b M 503

3 Hence congruent z-angles (Euclid I.27) yield that the lines AB and M a M b are parallel. The extensions of the already produced ray AM b intersect both parallel segments AB and M a M b 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 BM a intersect both parallel segments AB and M a M b at angles β. By the extended ASA congruence, the rays AM b and BM a do intersect, say at point C, and (8.3) A B C = Mb M a C Altogether, the congruences (1.11),(1.11) and (9.7) have produced four new congruent triangles. Furthermore M a,m b 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 <ABimplies 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 M c. By Pasch s axiom, the parallel intersects segment AC, sayin 504

4 Figure 8.3: Construction and properties of the midpoint triangle. point P b. Furthermore, the halved triangle AM c P b is equiangular to the original triangle ABC. By Proposition 8.2, all three sides of AM c P b are half of the corresponding sides of ABC. HenceAC =2AP b and P b = M b is the midpoint of side AC. Thismeansthat the parallel to BC through point M c does intersect triangle side AC in its midpoint. Similarly, one sees that the other two sides of the midpoint-triangle M a M b M c 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 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. 41 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. 505

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, AM b M c and ABC.

The triangles SM b M c and SBC are equiangular. We can apply the Double ASA Proposition 8.1 to the triangle SM b M c and the (longer) segment BC. Hence we conclude that SB =2SM b and SC =2SM c.

5 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, AM b M c and ABC. We see that they are equiangular, and that the side BC = 2M b M c. Hence the segment M b M c is parallel to the side BC. At first, we define point S as intersection of the two medians BM b and CM c. The triangles SM b M c and SBC are equiangular. We can apply the Double ASA Proposition 8.1 to the triangle SM b M c and the (longer) segment BC. Hence we conclude that SB =2SM b and SC =2SM c. By the fact of dividing the median BM b as 2 : 1, the point S on median BM b is uniquely defined. Now define point T as the intersection point of medians BM b and AM a. The same argument as above shows that point T divides the median BM b 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 A 0 B 0 C 0 with all three sides doubled. The original triangle ABC is congruent to the midpoint triangle A m B m C m of A 0 B 0 C 0. ABC = A m B m C m The altitudes of the triangle A m B m C m are the perpendicular bisectors of the sides of the larger triangle A 0 B 0 C 0. Hence they intersect in one point. This point is the circum-center of A 0 B 0 C 0, as well as the orthocenter of A m B m C m. Proposition 8.5. In Euclidean geometry, every triangle ABC is the midpoint triangle of a larger triangle A 0 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 506

6 Figure 8.5: Every triangle ABC is the midpoint triangle of a larger triangle A 0 B 0 C 0. lying on opposite sides of AC. Similarly, are lying on opposite sides of AC. Finally CB 0 A = ABC A 0 CB = ABC are lying on opposite sides of BC. Because the angle sum is α + β + γ =2R, pointc lies on segment A 0 B 0. Furthermore B 0 C = AB = CA 0. Hence C is the midpoint of segment A 0 B 0. Similarly, we see that A is the midpoint of segment B 0 C 0 and B is the midpoint of segment C 0 A 0. hence the given triangle ABC is the midpoint-triangle of the larger triangle A 0 B 0 C 0. Proof II: Erect the perpendiculars on the altitudes. One constructs the doubled triangle A 0 B 0 C 0 by erecting the perpendiculars on its altitudes at all three vertices of triangle ABC. Let p a,p b,p b be these three perpendiculars. Using angle sums, one can check that (p a, AC) = γ and (p c, CA) = α By the extended ASA congruence, lines p a and p c 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 507

7 These two triangles lie on opposite sides of AC. Similarly, one get a new triangle A 0 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 p a,p b,p b yield a larger triangle A 0 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 A 0 B 0 C 0 has angles smaller thantheanglesofthegiven 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 A 0 B 0 C 0. By Proposition 9.1, the altitudes of the triangle ABC are the perpendicular bisectors of the sides of the larger triangle A 0 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 A 0 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 e A and e B 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 508

8 intersect, say in point I c. 42 Now one argues similarly as in the case of the in-circle. Point I c has congruent distances to all three sides of the triangles, and hence lies on the interior bisector of angle BCA, too. PointI c 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 F a,f b and F c denote the foot points of the three altitudes. The triangle F a F b F c 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. 42 The unique parallel to e A through point B is different to e B. 509

The case of an acute triangle. We shall repeatedly use the congruence of circumference angle, stated by Euclid III.21. Both angle CAF a CBF b = R γ because of the angle sum and the right angles.

9 The case of an acute triangle. We shall repeatedly use the congruence of circumference angle, stated by Euclid III.21. Both angle CAF a CBF b = R γ because of the angle sum and the right angles. By the converse Thales theorem, the four points C, F b,f c and B lie on a semicircle with diameter BC. Hence the congruence of circumference angles (Euclid III.21) implies that CBF b = CFc F b. Similarly, the four points C, F a,f c and A lie on a semicircle with diameter AC, and the congruence of circumference angles implies CAF a = CFc F a. Hence CF c F a = CFc F b, and the altitude CF c bisects the angle F a F c F b 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 A 1 B 1 C 1, and one obtuse A 2 B 2 C 2. Proof. The first good exercise. Do it! 510

10 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 toapointa such that the three points A, A and C lie on a line and A C = CA. Ifk>0, the center C lies outside the segment AA.Ifk<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 M a,m b and M c 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 H 2 of the midpoint triangle is mapped to the orthocenter H of triangle ABC. By Proposition 9.1, the orthocenter H 2 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. 511

11 Figure 8.8: The Euler line comes from a dilation with center S. 512

Problem 2.1. Complete the following proof of Euclid III.20, referring to the figure on page 1.

Problem 2.1. Complete the following proof of Euclid III.20, referring to the figure on page 1. Math 3181 Dr. Franz Rothe October 30, 2015 All3181\3181_fall15t2.tex 2 Solution of Test Name: Figure 1: Central and circumference angle of a circular arc, both obtained as differences Problem 2.1. Complete