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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
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
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
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,
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
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!).
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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!
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