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Transcript

5 5.1 Hyperbolic Triangle Geometry Hyperbolic trigonometry Figure 5.1: The trigonometry of the right triangle. Theorem 5.1 (The hyperbolic right triangle). The sides and angles of any hyperbolic right triangle with γ the right angle satisfy (5.1) (5.2) (5.3) (5.4) (5.5) sinh a and sinh c tanh b and cos α = tanh c cosh c = cosh a cosh b cos α cosh a = and sin β cosh c = cot α cot β sin α = sinh b sinh c sinh a cos β= sinh c sin β = cosh b = cos β sin α Proof of the formula (5.2) for the cos of an angle. The formula has to be derived from the geometry in the Poincaré and Klein model, which are best used together. 827 We depict the given right triangle ABC in the Poincaré model with vertex A = O at the center. The opposite side is depicted as a circle around a⊥ . Let the transformation to the Klein model produce triangle ALK. Still, the right angle at vertex C is transformed to a right angle at K, since AK is a diameter. We use the underlying Euclidean geometry for the ALK and obtain cos α = tanh b |OK| = |OL| tanh c We have used that the hyperbolic distance to the center in the Klein model is given by b = s(O, K) = 1 1 + |OK| ln = tanh−1 |OK| 2 1 − |OK| Hence |OK| = tanh b, and similarly |OL| = tanh c. Proof of the formula (5.1) for the sin of an angle. We use the same setting as in the proof given above. Once more, the given right triangle ABC is depicted with vertex A = O at the center. In the Poincaré disk model, the opposite side is depicted as a circle around a⊥ = K . The translation to Klein’s model produces the points C → K and B → L, as well as for their images C → K and B → L by circular inversion. The underlying Euclidean geometry for the K L B yields sin β = |BL | |BB | |BL | = = |BK | |CK | |CC | We use now the hyperbolic distance to the center in the Poincaré model: b = s(O, C) = ln 1 + |OC| = 2 tanh−1 |OC| 1 − |OC| Hence |OC| = tanh 2b and |CC | = |OC | − |OC| = Similarly we get |BB | = 2 . sinh c 2 sinh(b/2) cosh(b/2) − = cosh(b/2) sinh(b/2) sinh b Thus the required sin is sin β = |BB | sinh b = |CC | sinh c All remaining formulas can be derived using only trigonometric identities. 828 Proof of the Pythagorean formula (5.3). Because of the Pythagorean theorem in its Euclidean form, we know that tanh2 b sinh2 a + tanh2 c sinh2 c sinh2 c · cosh2 b = sinh2 b · cosh2 c + sinh2 a · cosh2 b 1 = cos2 α + sin2 α = (1 + sinh2 c) · cosh2 b = sinh2 b · cosh2 c + (1 + sinh2 a) · cosh2 b cosh2 c · cosh2 b − sinh2 b · cosh2 c = cosh2 a · cosh2 b cosh2 c = cosh2 a · cosh2 b cosh c = cosh a · cosh b A side in terms of two angles (5.4) . We get from the hyperbolic Pythagorean theorem cosh a = tanh b sinh b cos α cosh c = : = cosh b tanh c sinh c sin β Similarly, we get cosh b = cos β sin α Formula (5.5) now follows immediately from the hyperbolic Pythagorean theorem (5.3). Problem 5.1 (The hyperbolic area of circle, calculated in Archimedes fashion). We use a regular n-gon with radius of circum-circle R to get the area of the circle from the of the defect, taking the limit n → ∞. (a) We calculate cot α for a right triangle with hypothenuse R and β = πn . (b) From the defect δ = π2 − α − β of one of 2n right triangles, we add up the total defect for the n-gon to be π − α − 2π Δn = 2n 2 (c) We need the ﬁrst term in the expansions a0 + an1 + na22 + . . . for the quantities π (5.6) − α = cot α = tan 2 π (5.7) −α= 2 Δn = (5.8) 829 From the last expansion, we calculate limn→∞ Δn , which is the area of the circle with radius R. Answer. (a) The regular n-gon consists of n nonoverlapping isosceles triangles. We drop the perpendiculars from the center O and get 2n right triangles. Each one of them has the angle β = πn at the center and the hypothenuse R. From the trigonometry of the right triangle we get cot α cot β = cosh R π cot α = (cosh R) tan n (c) We need the ﬁrst term in the expansions a0 + (5.9) tan (5.10) a1 n + a2 n2 + . . . for the quantities π π − α = cot α = (cosh R) tan = (cosh R) + O(n−3 ) 2 n n π π π −3 − α = arctan (cosh R) + O(n ) = (cosh R) + O(n−3 ) 2 n n π (5.11) Δn = 2n π 2 − α − 2π = 2π(cosh R − 1) + O(n−2 ) lim Δn = 2π(cosh R − 1) = 4π sinh2 (5.12) n→∞ R 2 which is the area of the circle with radius R. 5.2 The orthocenter Theorem 5.2 (The conditional orthocenter—hyperbolic version). The three altitudes of an acute or right triangle always intersect in one point. For an obtuse triangle, the three altitudes may intersect or not. There are three possible cases: (i) The three altitudes of an obtuse triangle intersect in one point. (ii) The three altitudes of an obtuse triangle are all divergent parallel to each other. There exists a line p perpendicular to all three altitudes. (iii) The three altitudes of an obtuse triangle are all asymptotically parallel to each other. Corollary 60. point. 60 If any two altitudes intersect, then all three altitudes intersect in one 60 The ﬁrst two sentences of the Corollary are stated in neutral geometry. A proof of that part in neutral geometry is clearly valid a real—but not Klein-bottle—of wine. 830 If the altitudes of two sides of a triangle have a common perpendicular, then the altitudes of all three sides have a common perpendicular. If the altitudes of two sides of a triangle are asymptotically parallel, then the altitudes of all three sides are asymptotically parallel. Figure 5.2: If any two altitudes intersect, one can put the intersection point at the center of Klein’s disk. Three drawings are given: for an acute, right, or obtuse triangle. We need to clarify some terms about the use of any mathematical models, as Klein’s or Poincaré’s: Deﬁnition 5.1. A theorem or a feature of a ﬁgure is part of neutral geometry if and only if it can be deducted assuming only the axioms of incidence, order, congruence. The facts of neutral geometry are valid in both Euclidean and hyperbolic geometry— as well as the more exotic non-Archimedean geometries. Deﬁnition 5.2. A feature of a ﬁgure drawn inside Klein’s model (as for example an angle, midpoint, altitude or bisector) is called absolute if it appears in the same way both for the underlying Euclidean plane, on which the model is based, and the hyperbolic geometry inside the model. Remark. Here are some features that appear absolute, both as features of hyperbolic geometry and in the underlying Euclidean plane: An angle with the center of Klein’s disk appears as an absolute angle. A right angle of which one side is a diameter appears absolute. A perpendicular bisector or an angle bisector which is a diameter appears absolute. Proof, using Klein’s model of hyperbolic geometry. As stated in Proposition 9.5, any two altitudes of an acute or right triangle do intersect. It can, but does not need to happen that the altitudes of an obtuse triangle intersect. These are simple facts in neutral geometry. If any two altitudes of the triangle intersect, we obtain an easily understandable picture in Klein’s model: just put the intersection point H into the center of Klein’s 831 disk. In the ﬁgure on page 831, I have given illustrations for an acute, right and an obtuse triangle. Why does that picture come out that simple? Here are the logical steps for the reason: The right angles at the foot points of the two intersecting altitudes appear undistorted as absolute right angles. We can now apply Theorem 8.3 from Euclidean geometry to the Euclidean triangle in the underlying Euclidean plane of Klein’s model. Hence the third Euclidean altitude passes through the intersection point of the two other ones, which we have put at the center O. Because the right angle at the foot point of the third altitude is depicted undistorted, too, all three altitudes are both Euclidean as well hyperbolic altitudes. Hence the three hyperbolic altitudes intersect at the center H. For an obtuse triangle, two new cases (ii) and (iii) do occur in hyperbolic geometry. They are illustrated in the ﬁgure on page 832. Of course, one can no longer put the orthocenter in the center of Klein’s disk, because it does not exist! Instead, we put the Figure 5.3: The altitudes of an obtuse triangle may or may not intersect. A convenient drawing puts the vertex with the obtuse angle at the center of Klein’s disk. vertex C with the obtuse angle into the center of the disk. In that way, still the three right angles at the foot points of the altitudes are depicted as absolute right angles. All three altitudes are both Euclidean as well as hyperbolic altitudes. Depending where the altitudes intersect, we get the three cases: (i) The intersection point H lies inside Klein’s disk. The three altitudes intersect in one point. (ii) The intersection point H lies outside Klein’s disk. The three altitudes are all divergent parallel to each other. There exists a line l perpendicular to all three altitudes. (iii) The intersection point H lies on the boundary of Klein’s disk. The three altitudes are all asymptotically parallel to each other. 832 In the ”genuine hyperbolic case” (ii), point H lies outside Klein’s disk. Hence H is not a point of the hyperbolic plane, but a so-called ultra-ideal point. The polar of point H yields a common perpendicular l to the three altitudes. Recall that the boundary circle δD of Klein’s disk is called the circle of inﬁnity. Finally, we get the ”borderline case” (iii) if point H lies on the circle of inﬁnity. Neither two of the three altitudes intersect, nor do any two of them have a common perpendicular. 5.3 About the circum-circle We begin by recalling Propositions 9.2 and 9.3 from neutral triangle geometry. Theorem 5.3 (The perpendicular bisectors and their meaning). Given any triangle, three diﬀerent cases can occur. ”As seen in Euclidean geometry” The three perpendicular bisectors intersect in one point. The triangle has a circum-circle, and the intersection point is the center of the circum-circle. This case always occurs for an acute or right midpoint-triangle. Too, it can—but does not need to occur—if the midpoint-triangle is obtuse. The orthocenter of the midpoint triangle Ma Mb Mc , is the circum-center of the larger original triangle ABC, too. (H2O) H2 = O ”The genuine hyperbolic case” The three bisectors are all divergently parallel to each other. There exists a line l perpendicular to all three bisectors. All three vertices have the same distance from line l. Hence there exists an equidistance line through the three vertices of the triangle. ”The borderline case” The three perpendicular bisectors are all asymptotically parallel to each other. Neither two of the three bisectors intersect, nor do any two of them have a common perpendicular. The three vertices lie neither on a circle nor an equidistance line. There exists a horocycle through the three vertices of the triangle. In hyperbolic geometry, all three cases do occur, whereas Euclidean geometry leads always to the ﬁrst case. Corollary 61. Any two perpendicular bisectors of sides of a triangle intersect iﬀ the bisectors of all three sides intersect in one point iﬀ the triangle has a circum-circle. The perpendicular bisectors of any two sides of a triangle have a common perpendicular iﬀ the perpendicular bisectors of all three sides have a common perpendicular iﬀ the three vertices lie on an equidistance line. The perpendicular bisectors of two sides of a triangle are asymptotically parallel iﬀ the perpendicular bisectors of all three sides are asymptotically parallel iﬀ the three vertices lie on a horocycle. 833 Figure 5.4: The bisectors and altitudes of the midpoint triangle inside Klein’s disk model. For a triangle with circum-center, three drawings are given: for an acute, right, or obtuse midpoint triangle. Proof for hyperbolic geometry, using Klein’s model. Most statements just repeat Proposition 9.6 from neutral geometry. As a new feature in hyperbolic geometry, we can use Klein’s model. Thus we can conﬁrm that all three mentioned cases actually do occur. Too, the meaning of the ”borderline case” is clariﬁed. By Corollary 24, the bisectors of the sides of the original triangle ABC are the altitudes of the midpoint-triangle Ma Mb Mc . Thus they form six right angles, in neutral geometry. We start with the case that any two altitudes of the midpoint-triangle do intersect. By Corollary 24, the altitudes of the midpoint-triangle are side bisectors for the original triangle. Now Proposition 9.2 implies that all three side bisectors intersect in one point. This point is the circum-center of the original triangle and the orthocenter of the midpoint-triangle as stated in formula (H2O). We have arrived at the case ”As seen in Euclidean geometry”, for which the triangle has a circum-circle. The case ”As seen in Euclidean geometry” always occurs for an acute or right midpoint-triangle, because any two altitudes of an acute or right triangle do intersect by Proposition 9.5. It can, but does not need to happen that the altitudes of an obtuse midpoint-triangle intersect. In Klein’s model an easily understandable picture is obtained by putting point O into the center of Klein’s disk. In the ﬁgure on page 834, I have given illustrations for an acute, right and an obtuse midpoint-triangle. Remark. For an acute midpoint-triangle, both vertex C and circum-center O lie on the same side of the longer triangle side AB. For an obtuse midpoint-triangle, vertex C and circum-center O lie opposite sides of the longer triangle side AB. But, indeed, the other two cases already mentioned in Proposition 9.6 do occur in hyperbolic geometry, too. They are illustrated in the ﬁgure on page 835. Of course, one can no longer put the circum-center in the center of Klein’s disk, because it does not exist! Instead, we put the midpoint Mc of the longest side in the center of the 834 Figure 5.5: The altitudes of the midpoint triangle intersect outside Klein’s disk. disk. Of the six right angles between the perpendicular bisectors and the sides of the triangles ABC and Ma Mb Mc , only four are depicted undistorted. Because the three right angles between the bisectors and the sides of the midpoint-triangle Ma Mb Mc are still undistorted, it is straightforward to construct its virtual orthocenter H2 . In the ”genuine hyperbolic case”, point H2 lies outside Klein’s disk. Hence H2 is not a point of the hyperbolic plane. Instead, the polar of the ultra-ideal point H2 yields a common perpendicular l to the three altitudes of triangle Ma Mb Mc . But—since Proposition 24 is a theorem of neutral geometry—it is still true that the altitudes of the midpoint-triangle are the bisectors of the original triangle. Therefore line l is a common perpendicular to the three bisectors of triangle ABC. Now still we drop the perpendiculars from the three vertices A, B and C onto line l. We end up with six lines that form right angles with line l! Too, we have reconstructed the situation from Example 9.1. The three points X, Y and Z are the intersections of l with the perpendicular bisectors. Once more, we get three Saccheri quadrilateral Y XAB and Y ZCB and ZXAC. Hence the three vertices A, B and C have congruent distances to line l, as to be shown. Recall that the boundary circle δD of Klein’s disk is called the circle of inﬁnity. Finally, we get the ”borderline case” if point H2 lies on the circle of inﬁnity. In that case, the three vertices lie neither on a circle nor an equidistance line. Neither two of the three bisectors intersect, nor do any two of them have a common perpendicular. 835 Figure 5.6: The three vertices A, B and C have congruent distances from the baseline l, which is chosen to be the horizontal diameter. The second drawing erases the part of the construction done outside the Poincaré disk. Problem 5.2. In the case that the three vertices of a triangle lie on an equidistance line, they have congruent distances to the baseline. In the ﬁgure of page 835, we get AX ∼ = BY ∼ = CY , where X, Y, Z are the foot points of the perpendiculars, dropped from the vertices onto the baseline l. This is hard to believe, especially because segment CZ intersects the opposite side AB of the triangle ABC, but the other two segments AX and BY do not intersect any side of the triangle. Redraw the ﬁgure in Poincaré’s model. The visible divergence of parallel lines makes existence of a baseline l more intuitive. Problem 5.3. In Klein’s disk model, there is given a triangle ABC with midpoint Mc of side AB at the center. Construct the midpoint-triangle, the six right angles and decide which case of Theorem 5.3 did occur. Problem 5.4. Given given is a triangle ABC with midpoint Mc of side AB at the center of a Klein disk, and midpoint Mb already speciﬁed. Construct the circle of inﬁnity δD of Klein’s disk. Finally get the midpoint-triangle, the six right angles and decide which case of Theorem 5.3 did occur. 836 Figure 5.7: Construction of the perpendicular bisector pb and the point H2 = O. Figure 5.8: Construction of the circle of inﬁnity δD and the point H2 = O. 5.4 Thales’ Theorem in hyperbolic geometry As already mentioned in the introductory section, Thales’ theorem does not hold in hyperbolic geometry. Once it is known that the angle sum of a triangle is less than two right angles, it is easy to see that the angle in a semicircle is acute. Nevertheless, even in neutral geometry, we can get a nice statement about a triangle in a semicircle. Proposition 5.1. Let one side of a triangle be a diameter of a semicircle and the third 837 Figure 5.9: Construction of the circle of inﬁnity δD and the point H2 = O—in this example it turns out to be outside the disk! vertex lie on that semicircle. Then the third vertex and the midpoints of the three sides are a Lambert quadrilateral. Figure 5.10: A triangle in a semicircle produces a Lambert quadrilateral. For the proof, the extra line we need is—once again—the radius to the third vertex. Proof. Let AB be the diameter and let C be the third vertex on the semicircle. Let Ma Mb Mc be the midpoint-triangle. Because Mc has congruent distances from vertices 838 A and C, it lies on the bisector pb , as follows from Proposition 9.1. Thus angle ∠Mc Mb C is a right angle. By the same reasoning, congruent distances Mc B ∼ = Mc C imply that Mc lies on the bisector pa , and hence angle ∠Mc Ma C is right. The bisector pb bisects the angle ∠AMc C, and the bisector pa bisects the supplementary angle ∠BMc C. As shown in paragraph of in-circle and ex-circles, the bisectors of supplementary angles are perpendicular. Hence pa and pb are perpendicular, and hence angle ∠Ma Mc Mb is right. Thus we have checked that the quadrilateral CMa Mc Mb has three right angles, resulting in a Lambert quadrilateral. Corollary 62. In neutral geometry, the angle in a semicircle is a right angle if and only if a rectangle exists. Reason. If a rectangle exists, we know from the second Legendre theorem that every Lambert quadrilateral is a rectangle. Hence the Lambert quadrilateral constructed in Proposition 5.1 is a rectangle. The right angle at vertex C is now Thales’ right angle in a semicircle. Conversely, if the angle in a semicircle is right, the Lambert quadrilateral constructed in Proposition 5.1 is a rectangle, and hence a rectangle does exist. Remark. The corollary holds without assuming Archimedes’ axiom. Similarly as we did in the introductory section, one can ask what happens for the third vertex inside or outside the semicircle. Thus one is lead to a strengthened version, and a converse to Proposition 5.1. Proposition 5.2 (A neutral version of the strengthened Thales’ theorem). Let one side AB of a triangle be a diameter of a circle. If the third vertex C lies inside the circle, the midpoint-triangle has an obtuse angle at the center of the semicircle. The quadrilateral CMa Mc Mb has an obtuse angle at vertex Mc and two acute angles at vertices Ma and Mb . If the third vertex C lies outside the circle, the midpoint-triangle has an acute angle at the center of the semicircle. The quadrilateral CMa Mc Mb has an acute angle at vertex Mc and two obtuse angles at vertices Ma and Mb . Corollary 63. Let one side of a triangle be a diameter of a circle. The third vertex C lies on that circle if and only if the midpoint-triangle has a right angle at the center of the semicircle. Proof in Klein’s model of hyperbolic geometry. The drawings on page 840 use Klein’s model with midpoint Mc at the center of the disk. I did not draw the circle of inﬁnity ∂D. The drawings include the semicircle, the original triangle ABC, its midpointtriangle Ma Mb Mc . By Corollary 24, the altitudes of the midpoint triangle are the side 839 Figure 5.11: A triangle with third vertex inside the circle produces an obtuse angle of the midpoint-triangle at center Mc . In case of the third vertex outside the circle, an acute angle is produced. bisectors of the original triangle. Hence the altitude of the midpoint-triangle dropped from vertex Mb onto the opposite side Ma Mc is equal to the perpendicular bisector pb . It has the foot point F . An additional perpendicular is dropped from Mc onto the side AC, and has the foot point G. Begin by distinguishing the following cases: (i) The midpoint-triangle has an obtuse angle at vertex Mc . (ii) The midpoint-triangle has a right angle at vertex Mc . (iii) The midpoint-triangle has an acute angle at vertex Mc , but an obtuse or right angle at vertex Ma . (iv) The midpoint-triangle has an acute angle at vertex Mc , but an obtuse or right angle at vertex Mb . (v) The midpoint-triangle is acute. To ﬁnish the proof of the Corollary, assume that case (ii) occurs. Clearly the ”waterpoint” is H2 = O = F = Mc . The side bisector pb = Mb F H2 and the line GMc are equal, because both are the perpendicular to Mc Ma at point Mc = F . We get a Lambert quadrilateral CMa Mc Mb . Because Mc lies on the bisector pb , both points C and A have congruent distances to Mc . Hence point C lies on the circle with diameter AB. Next we consider case (i). By Proposition 9.5, the orthocenter H2 = O of the midpoint-triangle lies inside the vertical angle to the obtuse angle ∠Mb Mc Ma . 61 Hence H2 lies on the side of line AB opposite to the points Ma , Mb and C. 61 Clearly, the orthocenter can be an ideal or ultra-ideal point of Klein’s model. 840 Answer. Because the angle ∠Ma Mc Mb is obtuse, the foot point F lies on the extension of side Ma Mc of the midpoint-triangle, and hence Mc lies between F and Ma . Sorting again points below and above line AB, we conclude that F and H2 lie below line AB, whereas Ma , Mb and C lie above line AB. −−−→ From here, one can check that ray Mb Mc lies inside the right angle ∠CMb H2 . Hence −−→ G lies on the ray Mb C, and GC is shorter than AC. Finally Mc C is shorter than Mc A, which conﬁrms that C lies inside the circle with diameter AB. It is left to the reader to check that in all remaining cases (iii) through (v) the −−−→ following can be deducted: Ray Mb Mc lies inside the right angle ∠AMb H2 . Hence GC is longer than AC. Finally Mc C is longer than Mc A, which conﬁrms that C lies outside the circle with diameter AB. Problem 5.5. Given is a triangle ABC with midpoint Mc of longest side AB at the center of a Klein disk and vertex C in the upper half disk. Use Proposition 5.2, and prove that the hyperbolic midpoints Mb and Ma are below the apparent Euclidean midpoint if and only if the midpoint-triangle is obtuse. The hyperbolic midpoints Mb and Ma are above the apparent Euclidean midpoint if and only if the midpoint-triangle is acute. Partial solution. The hyperbolic midpoint Mb is equal to the apparent midpoint of segment AC if and only if Mb = G. By the neutral version of the strengthened Thales’ theorem, given in Proposition 5.2 above, this happens if and only if the angle Ma Mc Mb is right. The assertion follows by continuity, once is hold in at least one example for obtuse and acute triangles. This can be established for some extreme isosceles triangles. 5.5 The centroid Theorem 5.4 (The Centroid). The three medians of a triangle intersect in one point. Remark. The ratio in which the intersection point divides the medians can be diﬀerent from 2 : 1. Proof. I use Klein’s model of hyperbolic geometry, to given a proof for the hyperbolic case. Given is a triangle ABC. A convenient position is achieved by putting triangle side AB on the horizontal diameter with its midpoint Mc = O in the center of the disk. The midpoints of the two other sides of the triangle do not appear as absolute midpoints, because of the distortion of distances in Klein’s model. Nevertheless, we can take advantage of Theorem 9.1. The perpendicular bisector pc of one side AB is perpendicular to the line l through the midpoints Ma and Mb of the other two sides BC and CA. Because this perpendicular bisector pc is a vertical diameter of the disk D, the right angle between pc and l appears as an absolute right angle. Hence, in the sense of underlying Euclidean geometry, the lines AB and Ma Mb get parallel . We are just back to the ﬁgure studied in the construction 14.3 of a Euclidean 841 Figure 5.12: The three medians intersect in one point because the harmonic quadrilateral ABMb Ma has Euclidean parallel sides. parallel— in the section about constructions by restricted means. By the Theorem of the harmonic quadrilateral (see Theorem 14.1), the three lines CMc , AMa and BMb intersect in one point, as to be shown. 5.6 The in-circle, ex-circles, and the orthic triangle Problem 5.6 (Conditional ex-circle—hyperbolic version). Provide the details of the proof for the following proposition. Part is easy to verify in Klein’s model—once one knows how to interpretate Klein’s model. If any two of the exterior bisectors at vertices A and B, and the interior bisector of the third angle ∠BCA intersect, then all these three bisectors intersect in one point. In that case, the triangle has an ex-circle touching side AB from outside, and the extensions of the two other sides. If any two of these three bisectors have a common perpendicular, then three have a common perpendicular. Furthermore, there exists a unique equidistance line touching side AB from outside, and the extensions of the two other sides. If any two of these three bisectors are asymptotically parallel, then all three are asymptotically parallel. Furthermore, there exists a unique horocycle touching side AB from outside, and the extensions of the two other sides. Theorem 5.5. For an acute triangle, the altitudes are the inner angular bisectors of the orthic triangle. For an obtuse triangle, only the altitude dropped from the obtuse angle is an inner angular bisector. The other two altitudes are exterior angular bisectors of the orthic triangle. 842 Figure 5.13: The altitudes are angular bisectors of the orthic triangle. Proof. This is easy to verify in Klein’s model! One needs to choose the center of the disk diﬀerently for an acute and an obtuse triangle. For an acute triangle, I put the orthocenter H at the center of the disk. Again the altitudes as well as the angular bisectors and the in-circle of the orthic triangle become absolute. For an obtuse triangle, I put vertex C at the center of the disk. Still the right angles at the foot points of the three altitudes appear as absolute right angles. But only the altitude dropped from the obtuse angle is an inner angular bisector of the orthic triangle. The other two inner bisectors are just the extensions of the two sides of the triangle at the obtuse angle. The other two altitudes get exterior angular bisectors, since it is a fact of neutral geometry that interior an exterior angular bisectors are perpendicular to each other. Problem 5.7. Prove that the orthic triangle of an acute triangle has an in-circle and three ex-circles. The orthic triangle of an obtuse triangle has an in-circle and either two or three ex-circles. The orthic triangle has three ex-circles if and only if the original triangle has an orthocenter. Proof. For an acute triangle, one can put any vertex at the center of Klein’s disk, too. One sees that the center becomes the intersection of two exterior and one interior angular bisector, and one gets an ex-circle around that center, all depicted in Klein’s model absolutely. (Of course, one can only depict one of the four circles in question absolutely at a time!). 843 Figure 5.14: Centers for any ex-circle of the orthic triangle can only be the three vertices or the orthocenter of the original triangle. For an obtuse triangle, one can put the two vertices with acute angles at the center, and get two ex-circles of the orthic triangle. To get the third ex-circle, one needs to put the orthocenter at the center of Klein’s disk, which is possible if and only if the orthocenter exists. Figure 5.15: The triangle has no circum-circle—drawn in Klein’s model! 844