19 Advanced Euclidean Triangle Geometry

19.1 Definitions

Theorem 19.1 (The Euler line). The orthocenter, the centroid, and the circum-center of a triangle lie on the Euler line. The centroid trisects the segment joining theorthocenter and the circum center. In the exceptional case of an equilateral triangle, allthree 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 thetriangle. The circum-circle of the midpoint triangle �MaMbMc is called the Feuerbachcircle 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 ofa triangle, the foot points Fa, Fb, Fc of its altitudes, and the midpoints Ha, Hb, Hc of thesegments joining the orthocenter H (intersection point of the three altitudes) to the threevertices, 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 ortho-center and the circum-center.

The orthocenter H, the center N of the Feuerbach circle, the centroid S, and thecircum-center O have the order H,N, S,O on the Euler line. Furthermore,

6 · |NS| = 3 · |SO| = 2 · |HN | = 2 · |NO| = |HO|

and (NO,SH) are harmonic points.

Main Theorem 30 (Feuerbach’s Theorem). The Feuerbach circle touches the in-circle 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 lineof an equilateral triangle is not defined.

Problem 19.3. Prove that the Euler line of an isosceles—but not equilateral—triangleis 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 theopposite side. For a right triangle, these two points are just the orthocenter and thecircum center. Thus the centroid trisects the segment joining the orthocenter and thecircum 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 theorthocenter H = O = S—nothing else is claimed.

We now discard this special case and assume that S �= O. Let E be the point online SO such that |ES| = 2|SO|, with centroid S lying between E and O. Let M bethe midpoint of side AB and F be the foot point of the altitude dropped from vertexC onto side AB. Euler’s theorem follows easily from the

Lemma (**). Point E lies on the altitude FC.

Because of Claim (**), the point E lies on all three altitudes of the triangle. Henceit is the orthocenter E = H, and the three points S,O and H lie on a line and

|HS| = 2|SO|(19.1)

|HA| = 2|OMa| , |HB| = 2|OMb| , |HC| = 2|OMc|(19.2)

as to be shown.

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Remark. In this way, Claim (**) yields an independent proof that the three altitudesintersect in one point.

Generic case: Assume sides AC and BC are not congruent, and the angle at vertex Cis not a right angle.

Reason for Claim (**) in the generic case. By the centroid theorem, the centroid S di-vides the median CM at the ratio 2 : 1. By the construction above, the segment EO isdivided by S in the same ratio 2 : 1. Hence the two triangles

�SOM ∼ �SEC

are similar by Euclid VI.6. Hence

|EC| = 2|OM |(19.3)

By Euclid VI.5, the two triangles are equiangular, too. (One can independently getthese two statements from Double SAS, Proposition 8.4) Via congruent z-angles, oneconcludes that the two segments OM and EC are parallel. Because OM is perpendicularto the side AB, the parallel segment EC is perpendicular to AB, too. Hence point Elies 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 rightangle, then C = H and O = M , by Thales theorem and its converse. The Euler line isthe median CM . By the centroid theorem, the centroid S trisects the segment joiningthe vertex C = H and the midpoint M = O of the opposite side. Thus the centroidtrisects the segment joining the orthocenter and the circum center— as stated in Euler’stheorem.

Now assume that the triangle is isosceles with congruent sides AC ∼= BC. ClearlyM = F . The Euler line OS is the symmetry axis. But this is one altitude, too. Hencepoint E lies on the one altitude which is the symmetry axis. Too, one can check thatstatement (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 �NOM and �NHHc are used to get the nine-pointcircle.

Proof of the nine-point theorem. The notation and set-up from the proof Euler’s theo-rem is used, once more. Let N be the midpoint of segment OH on the Euler line. LetM be the midpoint of side AB and F be the foot point of the altitude dropped fromvertex C onto side AB. As a new feature, let Hc be the midpoint of segment HC. Thenine-point theorem follows easily from the

Lemma (*). The three points M,F and Hc lie on a circle N around N . SegmentMHc is a diameter of this circle. The radius of circle N is half of the radius of thecircum-circle of triangle �ABC.

Because of Claim (*), the circle N turns out to be the same, no matter whether itsdefinition 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 thegeneric case, which is as follows:

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Generic case: Assume sides AC and BC are not congruent, and the angle at vertex Cis not a right angle.

From Euler’s theorem and its proof, we know

|HS| = 2|SO|(19.4)

|HC| = 2|OM |(19.5)

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

�NOM ∼= �NHHc

are congruent by SAS congruence. Indeed

∠NOM ∼= ∠NHHc because these are z-angles for the parallel lines OM ‖ HHc = CF

OM ∼= HHc follows by (19.5) and the definition of Hc via 2OM ∼= HC ∼= 2HHc

NO ∼= NH because N is the midpoint of segment OH

Hence ∠ONM ∼= ∠HNHc. Furthermore, these are vertical angles, since N is themidpoint of segment OH, and points C and Hc lie on the opposite side of the Euler lineOH than point M .

Point N is the midpoint of segment MHc, since NM ∼= NHc. The angle ∠MFHc

is a right angle. Hence the converse Thales’ theorem shows that the three points M,Fand Hc lie on a circle, as claimed.

Finally, we check the radius of this circle. Because of result (19.5) and definitionof Hc, we get 2OM ∼= HC ∼= 2CHc. Thus MO and CHc are two parallel congruentsegments, and the quadrilateral �MOCHc is a parallelogram. Hence OC ∼= MHc =2NM . This confirms that the radius NM of circle N is half of the radius OC of thecircum-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 threepoints M,F and C lie on a circle with diameter MC, which is the radius of the circum-circle.

Now assume that the triangle is isosceles with congruent sides AC ∼= BC. ClearlyM = F . We check that N is the midpoint of segment MHc. One checks that 2|OHc| =|O2H| = 2|HM | and hence OHc

∼= HM and NHc∼= NM .

19.4 Proof of Feuerbach’s Theorem

Proposition 19.1 (Angle between the Feuerbach circle and the sides of thetriangle). 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 sideCA 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 Tbe on t on the side of AB opposite to C. Here is the case α < β, as in the drawing.

∠TMcFc∼= ∠TMcMa − ∠FcMcMa

∼= ∠McMbMa − ∠BMcMa

because, by Euclid III.32, the circumference angle (here of arc McMa) is congruent tothe angle between the chord and the tangent at its endpoint.

∼= ∠MaBMc − ∠BAC

because opposite angles in a parallelogram are congruent, and because z-angles betweenparallels are congruent. Finally

∠TMcFc∼= ∠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 ex-circle Ic touching side AB. Let Da, Db and Dc be the points where the in-circle touchestriangle sides a, b and c, respectively. Let Ea and Eb be the points on the extensions oftriangle sides a and b where the ex-circle Ic touches the extended sides, and let Ec bethe point where ex-circle Ic touches the side AB. Considering segment AB the base ofthe 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 theex-circle Ic in terms of the triangle sides a, b, c and the half perimeter s :=

a+b+c2. Show

that M is the midpoint of segment DE.

Answer. The segments on the two tangents to a circle from a point outside to thetouching points are congruent. Hence x = |ADb| = |ADc|, y = |BDc| = |BDa|, andz = |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 thepositions 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 in-circle I and the same ex-circle Ic. Note that the other three common tangents are justthe 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 intersectionpoint of the triangle side AB with the segment IIc. Points D′ and E ′ are the pointswhere 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 isβ − α.

Problem 19.7. Calculate the angle ∠D′GB between the two inner common tangents ofin-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

∠CGB =γ

2+ α

∠CGA =γ

2+ β

Line GC bisects the angles between the two inner tangents, hence ∠CGD′ ∼= ∠CGAand

∠CGD′ ∼= ∠CGA =γ

2+ β

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 tangentto the Feuerbach circle.

Proposition 19.2. The fourth tangent tc and the tangent t to the Feuerbach circle atmidpoint 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 Vdivide segment XY from inside and outside in the same ratio. In other words, the crossratio

(XY,UV ) =XU · Y VY U ·XV = −1

for the directed segments.

Figure 19.7: The points on Apollonius circle have all same ratio of distances from C andD.

On an angular bisector from vertex C, we mark the intersection point G with theopposite 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 segmentCG. The points on this circle have all same ratio of distances

|CA||GA| =

|CI||GI| =

|CIc||GIc|

from C and G. Especially, vertex C, the intersection point G of the angular bisectorwith the opposite side, and the centers I of in-circle and ex-circle Ic, are four harmonicpoints:

(CG, IIc) =CI ·GIGIc · CIc

= −1

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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 in-circle and all three ex-circles.

Reason. It is enough to show that the Feuerbach circle touches the in-circle I and theex-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

Mc �→ ∞ , D �→ D , E �→ E , G �→ F ,(19.6)

since the points of δ are mapped to themselves, and because (DE,FG) are harmonicpoints.

Question. Why are the four points (FG,DE) harmonic?

Answer. The four points F,G,D,E are the foot points of the perpendiculars onto lineAB 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 areorthogonal 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 ismapped to a line or circle. The second inner tangent tc is mapped to a circle which Ihave 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 totouching generalized circles. The inner tangent tc touches both the in-circle and the ex-circle Ic. Applying circular inversion by δ, we conclude that the corresponding imagestouch, too. Hence the circle φ touches the images of in-circle and ex-circle, which areagain 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 Feuerbachcircle?

Answer. I show that circle φ and the Feuerbach circle have points F and Mc, and thetangent at pointMc in common. This implies that the two circles are identical. Becausetc 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 withδ lie on φ, because they are fixed under inversion. Hence the tangent to φ at pointMc is parallel to tc. By a result above, the two lines tc and t are parallel, too. Hencecircle φ and the Feuerbach circle have the same tangent t at point Mc. Furthermoreboth 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 altitudedropped from vertex C. Circle φ goes through point F , because line tc goes throughpoint 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 itsthirteen 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 ofa triangle. That is clear, because any triangle can be mapped by a composition of trans-lation, rotation and dilation into one having the prescribed Euler line and Feuerbachcircle.

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 solutionsan entire set of rotated equilateral triangles.

If the orthocenter lies outside the Feuerbach circle, a solution exists only underthe following restrictions: The distance |NH| 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 thenine 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 getspoint C on the given ray. The foot points of the two other altitude Fa and Fb are theintersection points with the Feuerbach circle. Next, one draws the circle with diameterOC, and gets the midpoints of sidesMa andMb as intersection points with the Feuerbachcircle.

The second intersection of line HC with the Feuerbach circle yields the foot pointFc, and the line AB as perpendicular to the altitude hc at point Fc. Now one can use thecircum-center O and use the circum circle, since the center N of the Feuerbach circle isknow to be the midpoint of segment HO. One gets the vertices A and B as intersectionpoints 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 pointsC, Fa,Ma and B. On the triangle side b, one has the four points C, Fb,Mb and A. Onthe 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 theEuler line— leads to an acute triangle.

case (c) the orthocenter H inside the Feuerbach circle, and vertex C on the Eulerline— leads to an acute isosceles triangle.

case (c) the orthocenter H outside the Feuerbach circle, but no vertex on the Eulerline— leads to an obtuse triangle.

case (d) the orthocenter H outside the Feuerbach circle, and vertex C on the Eulerline— 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). Thethree foot points of the perpendiculars dropped onto the three sides of a triangle lie on aline 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 congru-ent 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 footpoints 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 thesituation as draw in figure 19.6, we now assume that point P lies on the arc between Band 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 ∼= ∠GFP are congruent.

The angles α = ∠CAB and ∠CPB = 180◦ − α1 − α2 add up to two right angles, byEuclid III.22. The quadrilateral figure �EPFG has at its vertices E,P, F the anglesα1, 180

◦−α1−α2 and α2—which add up to 180◦. Hence the quadrilateral is degeneratewith 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 ∠CPB < 180◦−α1−α2, and one gets a convex quadrilateral�EPFG.

If point P lies inside the circum circle, then ∠CPB > 180◦ − α1 − α2,and one getsa non convex quadrilateral �EPFG.

Problem 19.12. Assume that the point P is the intersection of the circum circle withthe 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 anextension of a second side.

(c) The triangle �EPF 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. Thecenter O of the circum circle is the intersection point of the perpendicular bisectors ofall three sides of the triangle. Especially, it lies on the perpendicular bisector m of sideBC. 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 circumcircle. Hence the two central angles ∠BOP and ∠COP are congruent. Hence the twocorresponding 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 α2 at

vertices A,B,C.F,E.

that line AP is the angular bisector of ∠BAC:

(*) ∠BAP ∼= ∠PAC ∼= α

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 �CFGP . Recall that it has acircum circle V , because of the two right angles ∠CFP ∼= ∠CGP = 90◦. Hence thetwo angles at the two opposite vertices at C and G add up to two right angles. Henceα2+ β + y + 90◦ = 180◦. We solve for angle y = ∠CGF and use the angle sum

α + β + γ = 180◦ to conclude

y = 90◦ − α

2− β =

γ − β2

as claimed in part (d).

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