Download 19 Advanced Euclidean Triangle Geometry

19
19.1
Advanced Euclidean Triangle Geometry
Definitions
Theorem 19.1 (The Euler line). The orthocenter, the centroid, and the circumcenter of a triangle lie on the Euler line. The centroid trisects the segment joining the
orthocenter and the circum center. In the exceptional case of an equilateral triangle, all
three centers are equal, but the Euler line is not defined.
Definition 19.1. Let Ma , Mb and Mc denote the midpoints of the three sides of the
triangle. The circum-circle of the midpoint triangle Ma Mb Mc is called the Feuerbach
circle or nine point circle .
These are the famous theorems about the Euler line and the Feuerbach circle:
Theorem 19.2 (The nine-point circle). The midpoints Ma , Mb , Mc of the sides of
a triangle, the foot points Fa , Fb , Fc of its altitudes, and the midpoints Ha , Hb , Hc of the
segments joining the orthocenter H (intersection point of the three altitudes) to the three
vertices, all lie on the nine-point circle, also called Feuerbach circle.
The center of the Feuerbach circle is the midpoint of the segment joining the orthocenter and the circum-center.
The orthocenter H, the center N of the Feuerbach circle, the centroid S, and the
circum-center O have the order H, N, S, O on the Euler line. Furthermore,
6 · |N S| = 3 · |SO| = 2 · |HN | = 2 · |N O| = |HO|
and (N O, SH) are harmonic points.
Main Theorem 30 (Feuerbach’s Theorem). The Feuerbach circle touches the incircle and all three ex-circles.
Problem 19.1. Find out which of these facts are easy to check for a right triangle.
19.2
Proof of Euler’s Theorem
Problem 19.2. Prove that S = O if and only if the triangle is equilateral.
Remark. For an equilateral triangle, it is easy to see that H = O = S. The Euler line
of an equilateral triangle is not defined.
Problem 19.3. Prove that the Euler line of an isosceles—but not equilateral—triangle
is its axis of symmetry.
Problem 19.4. Prove Euler’s theorem for a right triangle.
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Proof. If the angle at vertex C is a right angle, then C = H and O = M , by Thales’
theorem and its converse. The Euler line is the median CM . By the centroid theorem,
the centroid S trisects the segment joining the vertex C and the midpoint M of the
opposite side. For a right triangle, these two points are just the orthocenter and the
circum center. Thus the centroid trisects the segment joining the orthocenter and the
circum center— as stated in Euler’s theorem.
Figure 19.1: Similar triangles SOM ∼ SEC are used to prove Euler’s theorem.
Proof of Euler’s theorem. In the special case S = O, the triangle is equilateral, and the
orthocenter H = O = S—nothing else is claimed.
We now discard this special case and assume that S = O. Let E be the point on
line SO such that |ES| = 2|SO|, with centroid S lying between E and O. Let M be
the midpoint of side AB and F be the foot point of the altitude dropped from vertex
C onto side AB. Euler’s theorem follows easily from the
Lemma (**). Point E lies on the altitude F C.
Because of Claim (**), the point E lies on all three altitudes of the triangle. Hence
it is the orthocenter E = H, and the three points S, O and H lie on a line and
(19.1)
(19.2)
|HS| = 2|SO|
|HA| = 2|OMa | , |HB| = 2|OMb | , |HC| = 2|OMc |
as to be shown.
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Remark. In this way, Claim (**) yields an independent proof that the three altitudes
intersect in one point.
Generic case: Assume sides AC and BC are not congruent, and the angle at vertex C
is not a right angle.
Reason for Claim (**) in the generic case. By the centroid theorem, the centroid S divides the median CM at the ratio 2 : 1. By the construction above, the segment EO is
divided by S in the same ratio 2 : 1. Hence the two triangles
SOM ∼ SEC
are similar by Euclid VI.6. Hence
(19.3)
|EC| = 2|OM |
By Euclid VI.5, the two triangles are equiangular, too. (One can independently get
these two statements from Double SAS, Proposition 8.4) Via congruent z-angles, one
concludes that the two segments OM and EC are parallel. Because OM is perpendicular
to the side AB, the parallel segment EC is perpendicular to AB, too. Hence point E
lies on the altitude dropped from C onto side AB, as to be shown.
Reason for the Claim(**) in the exceptional cases. If the angle at vertex C is a right
angle, then C = H and O = M , by Thales theorem and its converse. The Euler line is
the median CM . By the centroid theorem, the centroid S trisects the segment joining
the vertex C = H and the midpoint M = O of the opposite side. Thus the centroid
trisects the segment joining the orthocenter and the circum center— as stated in Euler’s
theorem.
Now assume that the triangle is isosceles with congruent sides AC ∼
= BC. Clearly
M = F . The Euler line OS is the symmetry axis. But this is one altitude, too. Hence
point E lies on the one altitude which is the symmetry axis. Too, one can check that
statement (19.3) remains true.
19.3
Proof of the Nine-Point Theorem
Problem 19.5. Because of symmetry, some of the nine points
Ma , Mb , Mc midpoints of the sides
Fa , Fb , Fc foot points of altitudes, and
Ha , Hb , Hc midpoints of the segments joining the orthocenter H to the vertices
can become equal. How many points of them are left in the case of
(a) a generic right triangle with γ = R > α > β.
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(b) an isosceles acute triangle.
(c) an isosceles right triangle.
(d) an isosceles obtuse triangle.
(e) an equilateral triangle.
Figure 19.2: Congruent triangles N OM and N HHc are used to get the nine-point
circle.
Proof of the nine-point theorem. The notation and set-up from the proof Euler’s theorem is used, once more. Let N be the midpoint of segment OH on the Euler line. Let
M be the midpoint of side AB and F be the foot point of the altitude dropped from
vertex C onto side AB. As a new feature, let Hc be the midpoint of segment HC. The
nine-point theorem follows easily from the
Lemma (*). The three points M, F and Hc lie on a circle N around N . Segment
M Hc is a diameter of this circle. The radius of circle N is half of the radius of the
circum-circle of triangle ABC.
Because of Claim (*), the circle N turns out to be the same, no matter whether its
definition involves vertex A or B or C. Hence it is the nine-point circle.
Reason for Claim (*) in the generic case. Again, the main step is the proof for the
generic case, which is as follows:
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Generic case: Assume sides AC and BC are not congruent, and the angle at vertex C
is not a right angle.
From Euler’s theorem and its proof, we know
(19.4)
(19.5)
|HS| = 2|SO|
|HC| = 2|OM |
The orthocenter H is the intersection point of lines SO and CF . In the generic case,
OM CF are two different parallel lines. They are both perpendicular to AB.
Recall that Hc is the midpoint of segment HC. The two triangles
N OM ∼
= N HHc
are congruent by SAS congruence. Indeed
∠N OM ∼
= ∠N HHc because these are z-angles for the parallel lines OM HHc = CF
follows by (19.5) and the definition of Hc via 2OM ∼
OM ∼
= HC ∼
= HHc
= 2HHc
∼
NO = NH
because N is the midpoint of segment OH
Hence ∠ON M ∼
= ∠HN Hc . Furthermore, these are vertical angles, since N is the
midpoint of segment OH, and points C and Hc lie on the opposite side of the Euler line
OH than point M .
Point N is the midpoint of segment M Hc , since N M ∼
= N Hc . The angle ∠M F Hc
is a right angle. Hence the converse Thales’ theorem shows that the three points M, F
and Hc lie on a circle, as claimed.
Finally, we check the radius of this circle. Because of result (19.5) and definition
of Hc , we get 2OM ∼
= HC ∼
= 2CHc . Thus M O and CHc are two parallel congruent
segments, and the quadrilateral M OCHc is a parallelogram. Hence OC ∼
= M Hc =
2N M . This confirms that the radius N M of circle N is half of the radius OC of the
circum-circle of triangle ABC.
Reason for Claim (*) in the exceptional cases. If the angle at vertex C is a right angle,
then C = Hc = H and O = M . The converse Thales’ theorem shows that the three
points M, F and C lie on a circle with diameter M C, which is the radius of the circumcircle.
Now assume that the triangle is isosceles with congruent sides AC ∼
= BC. Clearly
M = F . We check that N is the midpoint of segment M Hc . One checks that 2|OHc | =
|O2 H| = 2|HM | and hence OHc ∼
= HM and N Hc ∼
= NM.
19.4
Proof of Feuerbach’s Theorem
Proposition 19.1 (Angle between the Feuerbach circle and the sides of the
triangle). The Feuerbach circle cuts the triangle side AB at point Mc at angle β − α.
Similarly, it cuts the triangle side BC at point Ma at angle γ − β, and the triangle side
CA at point Mb at angle α − γ.
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Figure 19.3: The angle between the Feuerbach circle triangle side AB is β − α.
Reason. Let t be the the tangent to the Feuerbach circle at point Mc , and let point T
be on t on the side of AB opposite to C. Here is the case α < β, as in the drawing.
∠T Mc Fc ∼
= ∠T Mc Ma − ∠Fc Mc Ma
∼
= ∠Mc Mb Ma − ∠BMc Ma
because, by Euclid III.32, the circumference angle (here of arc Mc Ma ) is congruent to
the angle between the chord and the tangent at its endpoint.
∼
= ∠Ma BMc − ∠BAC
because opposite angles in a parallelogram are congruent, and because z-angles between
parallels are congruent. Finally
∠T Mc Fc ∼
= ∠CBA − ∠BAC = β − α
The case α > β is almost similar, and left to the reader.
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Let I be the center of the inscribed circle or in-circle I, and Ic be the center of the excircle Ic touching side AB. Let Da , Db and Dc be the points where the in-circle touches
triangle sides a, b and c, respectively. Let Ea and Eb be the points on the extensions of
triangle sides a and b where the ex-circle Ic touches the extended sides, and let Ec be
the point where ex-circle Ic touches the side AB. Considering segment AB the base of
the triangle ABC, I write D = Dc , E = Ec and M = Mc for simplicity.
Figure 19.4: The touching points of the in-circle and one ex-circle.
Problem 19.6. Calculate the positions of the touching points for the in-circle, and the
ex-circle Ic in terms of the triangle sides a, b, c and the half perimeter s := a+b+c
. Show
2
that M is the midpoint of segment DE.
Answer. The segments on the two tangents to a circle from a point outside to the
touching points are congruent. Hence x = |ADb | = |ADc |, y = |BDc | = |BDa |, and
z = |CDa | = |CDb |. The three lengths x, y, z satisfy the equations
x+y
= c
y+z = a
x +z = b
which implies x + y + z = s, where
s :=
a+b+c
2
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is defined to be the half perimeter. Hence x = s − a, y = s − b, z = s − c. Because of
|AE| = |AEb |, |BE| = |BEa | and
a + |BEa | = b + |AEb | and
|BE| + |AE| = c
one concludes |AE| = y = s − b and |BE| = x = s − a. Thus we have calculated the
positions of the touching points for the in-circle, and the ex-circle. Together, we got
|AD| = |BE| = s − a. Hence M is the midpoint of DE as claimed.
Continuing the last problem, we draw the forth common tangent tc = D E of incircle I and the same ex-circle Ic . Note that the other three common tangents are just
the sides of ABC and their extensions. Let G be the intersection point of the two
”inner” common tangents tc and AB between the two circles. Point G is the intersection
point of the triangle side AB with the segment IIc . Points D and E are the points
where tc touches the in-circle and the ex-circle Ic .
Figure 19.5: The angle between the two inner common tangents of in-circle and ex-circle is
β − α.
Problem 19.7. Calculate the angle ∠D GB between the two inner common tangents of
in-circle and ex-circle.
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−→ −→
Answer. The ray CI = CIc is the inner angular bisector at vertex C. Euclid I.32 for an
exterior angle of the triangles AGC and CGB, yields
γ
+α
2
γ
∠CGA = + β
2
∠CGB =
Line GC bisects the angles between the two inner tangents, hence ∠CGD ∼
= ∠CGA
and
γ
∠CGD ∼
= ∠CGA = + β
2
Now angle subtraction at vertex G implies
∠D GB = ∠CGB − ∠CGD = α − β
as to be shown.
Figure 19.6: The common tangent of the in-circle and one ex-circle is parallel to the tangent
to the Feuerbach circle.
Proposition 19.2. The fourth tangent tc and the tangent t to the Feuerbach circle at
midpoint Mc are parallel.
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Definition 19.2. Four points X, Y, U, V are called harmonic if the two points U and V
divide segment XY from inside and outside in the same ratio. In other words, the cross
ratio
XU · Y V
(XY, U V ) =
= −1
Y U · XV
for the directed segments.
Figure 19.7: The points on Apollonius circle have all same ratio of distances from C and
D.
On an angular bisector from vertex C, we mark the intersection point G with the
opposite side, and the centers I of in-circle and ex-circle Ic .
Theorem 19.3. The circle with diameter IIc is indeed an Apollonius circle for segment
CG. The points on this circle have all same ratio of distances
|CI|
|CIc |
|CA|
=
=
|GA|
|GI|
|GIc |
from C and G. Especially, vertex C, the intersection point G of the angular bisector
with the opposite side, and the centers I of in-circle and ex-circle Ic , are four harmonic
points:
CI · GI
= −1
(CG, IIc ) =
GIc · CIc
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Figure 19.8: Inversion by circle δ maps I → I ,
Ic → Ic ,
AB → AB ,
tc → φ.
Main Theorem 31 (Feuerbach’s Theorem). The Feuerbach circle touches the incircle and all three ex-circles.
Reason. It is enough to show that the Feuerbach circle touches the in-circle I and the
ex-circle Ic . We use inversion by the circle δ with center Mc and diameter DE.
Question. Calculate the diameter of circle δ, assuming a > b.
Answer. In a problem above, we have shown |AD| = |BE| = s − a. Hence |DE| =
|AB| − |AD| − |BE| = c − 2(s − a) = c + 2a − (a + b + c) = a − b.
The circular inversion by δ maps the points
(19.6)
Mc → ∞ , D → D , E → E , G → F ,
since the points of δ are mapped to themselves, and because (DE, F G) are harmonic
points.
Question. Why are the four points (F G, DE) harmonic?
Answer. The four points F, G, D, E are the foot points of the perpendiculars onto line
AB from points C, G, I, Ic . Hence it is enough to show that these are harmonic points.
−→
−−→
But this is just evident from Apollonius’ theorem, because BI and BIc are two angular
bisectors of ∠CBG.
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The inversions maps the generalized circles
I → I ,
Ic → Ic ,
AB → AB ,
tc → φ
Indeed, the in-circle I and the ex-circle Ic are mapped to themselves, because they are
orthogonal to the inversion circle δ. The extended triangle side AB is mapped to itself,
because the center of inversion Mc lies on the line AB. Indeed any line or circle is
mapped to a line or circle. The second inner tangent tc is mapped to a circle which I
have defined to be φ.
Question. Why does the circle φ touch the in-circle I and the ex-circle Ic ?
Answer. Because of conservation of angles, touching generalized circles are mapped to
touching generalized circles. The inner tangent tc touches both the in-circle and the excircle Ic . Applying circular inversion by δ, we conclude that the corresponding images
touch, too. Hence the circle φ touches the images of in-circle and ex-circle, which are
again the in-circle I and the ex-circle Ic .
Question. The circle φ is defined as the inversion image of tc . Why is φ the Feuerbach
circle?
Answer. I show that circle φ and the Feuerbach circle have points F and Mc , and the
tangent at point Mc in common. This implies that the two circles are identical. Because
tc is a line, the inversion image φ goes through the center Mc of the inversion circle δ,
which is the inversion of the point ∞. Clearly, the two intersection points of tc with
δ lie on φ, because they are fixed under inversion. Hence the tangent to φ at point
Mc is parallel to tc . By a result above, the two lines tc and t are parallel, too. Hence
circle φ and the Feuerbach circle have the same tangent t at point Mc . Furthermore
both the Feuerbach circle and circle φ have the further point F in common. Indeed,
the Feuerbach circle goes through point F , because it is the foot point of the altitude
dropped from vertex C. Circle φ goes through point F , because line tc goes through
point G, and incidence is conserved by inversion.
19.5
Additional questions
Problem 19.8. Discuss the location of the Euler line and the Feuerbach circle with its
thirteen remarkable points in the case of
(a) a generic acute triangle with γ > α > β.
(b) a generic right triangle with γ = R > α > β.
(c) a generic obtuse triangle with γ > R > α > β.
(d) an isosceles acute triangle.
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(e) an isosceles right triangle.
(f ) an isosceles obtuse triangle.
(g) an equilateral triangle.
Given the Euler line and the Feuerbach circle, one can still not know any of the angles of
a triangle. That is clear, because any triangle can be mapped by a composition of translation, rotation and dilation into one having the prescribed Euler line and Feuerbach
circle.
Problem 19.9 (Construction problem). Construct a triangle from given Euler line,
−−→
Feuerbach circle, orthocenter H, and direction of the ray HC. Under which restrictions
does the problem has a solution? Is the solution unique?
Answer. If the orthocenter lies inside the Feuerbach circle and H = N , there always
−−→
exists a unique solution, once the direction of ray HC is specified up to an entire 360◦
turn. The solution is an acute triangle. In the special case H = N , one gets as solutions
an entire set of rotated equilateral triangles.
If the orthocenter lies outside the Feuerbach circle, a solution exists only under
the following restrictions: The distance |N H| needs to be less than the diameter of the
−−→
Feuerbach circle, and the ray HC needs to lie between the two tangents from orthocenter
H to the Feuerbach circle. Under this restriction, the solution is a unique obtuse triangle.
In the special case that H lies on the Feuerbach circle, one gets as solution a unique
right triangle.
Question. Explain the steps needed for the construction of the triangle ABC, and the
nine points on the Feuerbach circle.
−−→
Answer. The intersection of ray HC with the Feuerbach circle yields point Hc , the
midpoint of segment HC. One draws the circle with center Hc through H, and gets
point C on the given ray. The foot points of the two other altitude Fa and Fb are the
intersection points with the Feuerbach circle. Next, one draws the circle with diameter
OC, and gets the midpoints of sides Ma and Mb as intersection points with the Feuerbach
circle.
The second intersection of line HC with the Feuerbach circle yields the foot point
Fc , and the line AB as perpendicular to the altitude hc at point Fc . Now one can use the
circum-center O and use the circum circle, since the center N of the Feuerbach circle is
know to be the midpoint of segment HO. One gets the vertices A and B as intersection
points of line AB with the circum-circle.
The four remaining triangle sides and altitudes can be drawn with some redundancy,
which is good for accuracy. Indeed, on the triangle side a, one has the four points
C, Fa , Ma and B. On the triangle side b, one has the four points C, Fb , Mb and A. On
the altitude ha , one has three points A, H, Fa , and still gets Ha . On the altitude hb ,
one has three points B, H, Fb , and still gets Hb . Hence one has constructed the triangle
ABC and the nine points on the Feuerbach circle.
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Problem 19.10. Do the construction for several examples
(a) the orthocenter inside the Feuerbach circle, no vertex on the Euler line.
(b) the orthocenter inside the Feuerbach circle, and vertex C on the Euler line.
(c) the orthocenter outside the Feuerbach circle, but no vertex on the Euler line.
(d) the orthocenter outside the Feuerbach circle, and vertex C on the Euler line.
(e) the orthocenter on the Feuerbach circle.
To what type of triangle do these examples lead?
Answer. case (a) The orthocenter H inside the Feuerbach circle, but no vertex on the
Euler line— leads to an acute triangle.
case (c) the orthocenter H inside the Feuerbach circle, and vertex C on the Euler
line— leads to an acute isosceles triangle.
case (c) the orthocenter H outside the Feuerbach circle, but no vertex on the Euler
line— leads to an obtuse triangle.
case (d) the orthocenter H outside the Feuerbach circle, and vertex C on the Euler
line— leads to an obtuse isosceles triangle.
case (e) the orthocenter on the Feuerbach circle—leads to a right triangle.
Problem 19.11. Give reasons for the following conjectures:
Conjecture 1: The Euler line goes a vertex of a triangle if and only
if it is either isosceles or right.
Conjecture 2: The Euler line of an acute triangle cuts the shortest
and the longest sides.
Conjecture 3: The Euler line of an obtuse triangle cuts the two longest
sides, but not the shortest side.
Conjecture 4: For an obtuse triangle with γ obtuse, the two triangles
ABC and ABH have the same Feuerbach circle, but different
Euler lines.
Conjecture 5: All nine special points on the Feuerbach circle are
different, except in the cases of an isosceles, or a right triangle,
or an obtuse triangle for which the triangle with orthocenter
as one vertex, and two given vertices kept, is isosceles. (With
angle γ obtuse in ABC, the triangle ABH has the same Feuerbach
circle, and can be isosceles.)
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19.6
The Simson line
Proposition 19.3 (The Simson Line, first proved by W. Wallace, 1799). The
three foot points of the perpendiculars dropped onto the three sides of a triangle lie on a
line if and only if the point lies on the circum circle of the triangle.
Figure 19.9: By Euclid III.21, one gets congruent angles α1 at vertices A, B, E and congruent angles α2 at vertices A, C, F .
Proof. Let P be any point on the circum circle of ABC. Let E, F and G be the foot
points of the perpendiculars dropped from point P onto the lines AB, AC, and BC,
respectively. We claim that
(a) The four points A, P, E and F lie on a circle with diameter AP .
(b) The four points B, P, E and G lie on a circle U with diameter BP .
(c) The four points C, P, F and G lie on a circle V with diameter CP .
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Indeed this follows immediately, from the converse Thales theorem 32. To get the
situation as draw in figure 19.6, we now assume that point P lies on the arc between B
and C on which point A does not lie. From Euclid III.21, we can now conclude
(a) The three angles α1 = ∠CAP ∼
= ∠CBP ∼
= ∠GEP are congruent.
(b) The three angles α2 = ∠BAP ∼
= ∠BCP ∼
= ∠GF P are congruent.
The angles α = ∠CAB and ∠CP B = 180◦ − α1 − α2 add up to two right angles, by
Euclid III.22. The quadrilateral figure EP F G has at its vertices E, P, F the angles
α1 , 180◦ − α1 − α2 and α2 —which add up to 180◦ . Hence the quadrilateral is degenerate
with the three points E, F and G on one line. If point P lies outside the circum circle as
Figure 19.10: The Simson line is broken, if point P does not lie on the circum circle.
shown in figure 19.6, then ∠CP B < 180◦ − α1 − α2 , and one gets a convex quadrilateral
EP F G.
If point P lies inside the circum circle, then ∠CP B > 180◦ − α1 − α2 ,and one gets
a non convex quadrilateral EP F G.
Problem 19.12. Assume that the point P is the intersection of the circum circle with
the perpendicular bisector of side BC. Prove the following
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Figure 19.11: The Simson line is broken again.
(a) Point P lies on the bisector of the angle α = ∠BAC.
(b) One of the two points E and F lies on a side of ABC, and the other one of an
extension of a second side.
(c) The triangle EP F is isosceles.
(d) The angle between the triangle side a = BGC and its Simson line EGF is x :=
∠BGE = γ−β
.
2
Proof. Point P lies on the circum circle C of ABC, and hence outside of ABC. The
center O of the circum circle is the intersection point of the perpendicular bisectors of
all three sides of the triangle. Especially, it lies on the perpendicular bisector m of side
BC. Hence all three points O, G and P lie on m.
Furthermore G is the midpoint of side BC, and line m is a diameter of the circum
circle. Hence the two central angles ∠BOP and ∠COP are congruent. Hence the two
corresponding circumference angles ∠BAP and ∠CAP , are congruent, too. This implies
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Figure 19.12: The Simson line for a special case—one gets five congruent angles
vertices A, B, C.F, E.
α
2
at
that line AP is the angular bisector of ∠BAC:
(*)
α
∠BAP ∼
= ∠P AC ∼
=
2
Thus we have proved item (a). Part (b) and (c) are left to the reader.
Finally, we get the angle between and the triangle side a and its Simson line, claimed
in part (d). We use Euclid III.22 for the quadrilateral CF GP . Recall that it has a
circum circle V, because of the two right angles ∠CF P ∼
= ∠CGP = 90◦ . Hence the
two angles at the two opposite vertices at C and G add up to two right angles. Hence
α
+ β + y + 90◦ = 180◦ . We solve for angle y = ∠CGF and use the angle sum
2
α + β + γ = 180◦ to conclude
y = 90◦ −
γ−β
α
−β =
2
2
as claimed in part (d).
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