5. Orthic Triangle. Remarks There are several cyclic quadrilaterals

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1 5. Orthic Triangle. Let ABC be a triangle with altitudes AA, BB and CC. The altitudes are concurrent and meet at the orthocentre H (Figure 1). The triangle formed by the feet of the altitudes, A B C is the orthic triangle. Remarks There are several cyclic quadrilaterals : AC HB, BC HA, CA HB are cyclic. BCB C, ACA C, ABA B are cyclic. Figure 1: The sides of the orthic triangle are antiparallel with sides of the triangle ABC. We have A B is antiparallel to AB, B C is antiparallel to BC and C A is antiparallel to CA. Proposition 1 triangle A B C are If ABC is an acute triangle, then the angles of the 180 Â, 180 B and 180 Ĉ. Proof Since ACA C is cyclic, then C Â B = 180 C Â C = Â. Since ABA B is cyclic, then B Â C = 180 BÂB = Â. Thus C Â B = 180 Â. Similarly, for the other two angles of A B C. 1

2 Proposition The lengths of the sides of the orthic triangle are R sin(a) = a cos(a), R sin(b) = b cos(b) and R sin(c) = c cos(c), where R is the circumradius of the triangle ABC. Again, ABC is an acute triangle. Proof Since the points A B C lie on the ninepoint circle, the the circumcircle of A B C has circumradius R A B C which is one half of R. We now apply the sine rule to A B C. Then B C sin(â) = R A B C, B C and so sin(180 A) =. R, B C = R sin(â) = R sin(â) cos(â) = a cos(â). Remark In general, the side lengths of A B C are a cos(a), b cos(b) and c cos(c). Notation Proposition 3 If ABC is a triangle, we denote the area of ABC by S(ABC). The area of A B C is given by Proof S(A B C ) = R sin(â) sin( B) sin(ĉ). We have S(A B C ) = A C A B sin(â) = R sin( B) sin(ĉ) sin(â) = R sin(â) sin( B) sin(ĉ). Proposition 4 Let r A B C and R A B C denote the inradius and circumradius of the orthic triangle A B C. Then r A B C = R cos(â) cos( B) cos(ĉ) and R A B C = R.

3 Proof The value of R A B C follows from the fact that the ninepoint circle is the circumcircle of A B C and its radius is one half of the circumradius of ABC. For r A B C we have r A B C = = R S(A B C ) semiperimeter(a B C ) = (R /) sin(â) sin( B) sin(ĉ) (R/)(sin(Â) + sin( B) + sin(ĉ)) 8 sin(â) sin( B) sin(ĉ) cos(â) cos( B) cos(ĉ) 4 sin(â) sin( B) sin(ĉ) = R cos(â) cos( B) cos(ĉ). Proposition 5 If A B C is the orthic triangle of a triangle ABC and H is the orthocentre of ABC (Figure ), then (i) H is the incentre of A B C, and (ii) A, B and C are the centres of the excribed triangles. (i) Proof Since BA HC is cyclic, Since C Â H = C BH = A BH. CA HB is cyclic, HÂB = HĈB = HĈA. Since i.e. BCB C is cyclic, C BB = C ĈB A BH = HĈA. Figure : Thus C Â H = HÂB i.e. AA is the bisector of the angle at A. Similarly, BB and CC bisect the angles at B and C. Thus the point H is the incentre of the triangles A B C. 3

4 (ii) Since C A and B A are perpendicular to the internal bisectors C H and B H, then the point A is where the external angle bisectors meet. Furthermore, A lies on the internal bisector HA of the angle at A. Thus A is the centre of the escribed circle of A B C which is externally tangent to the side B C. Similarly for the other two vertices B and C. Theorem 1 (Haghal) The perpendiculars from the vertices A, B and C to the sides B C, C A and A B are concurrent at the circumcentre O of the triangle ABC. Proof Let T A be tangent to the circumcircle of ABC at the point A (Figure 3). We have that B C is antiparallel to the side BC and AT is antiparallel to BC (Step 1 of Feuerbach Theorem). Thus T A is parallel to BC. If O is the circumcentre of circumcircle of ABC, then AT is perpendicular to AO. Thus B C is perpendicular to AO. Similarly show that BO is perpendicular to A C and CO is perpendicular to A B. Figure 3: Theorem Among all inscribed triangles in a triangle ABC, the perimeter is minimized by the orthic triangle. Figure 4: 4

5 Proof Let A B C be inscribed in the triangle ABC (Figure 4). Let A be the reflection of A in the side AB and A be the reflection of A on the side AC (Figure 5). Then C A = C A and B A = B A Then if P denotes the perimeter, we have P(A B C ) = A B + B C + C A = B A + B C + C A = A B + B C + C A A A. Now consider the triangle A AA. We have AA = AA, AA = AA, so AA = AA. Figure 5: We also have A ÂB = A ÂB and A ÂC = A ÂC. Thus A ÂA = Â. Let γ be the angle AÂ A = AÂ A. If X is the point of point of intersection of the lines AA and A A, then A X A A = cos(γ). Thus A A = A X = cos(γ) A A = sin(â) A A, since 180 = γ +Â so γ +Â = 90 and thus cos(γ) = cos(90 Â) = sin(â). But A A = AA AA, so A A sin(â) AA 3. Thus, if A B C is an inscribed triangle, with B and C fixed, perimeter is minimised if A is the point A. Similarly the perimeter is further mimimised by taking B and C to be the points B and C respectively. Result follows. 5

6 Theorem 3 If ABC is an acute triangle which is not isosceles and A B C is the orthic triangle then the points A, B and C, where the sides B C and BC intersect, A C and AC intersect and A B and AB intersect, respectively, are collinear (Figure 6). Remark The line containing these points is called the orthic line of the triangle ABC. Proof If we are given two nonconcentric circles then the locus of points whose powers with respect to the circles is a line perpendicular to the line joining the centres of the circles. It is called the radical axis of the circles. Since BC B C is cyclic, then A B. A C = A C A B. If C is the circumcircle of ABC and C 9 ninepoint circle, thus ρ C (A ) = ρ C9 (A ). is the Figure 6: Similarly, ρ C (B ) = ρ C9 (B ) and ρ C (C ) = ρ C9 (C ). Thus the 3 points A, B and C lie on the radical axis of the circles C and C 9. 6