Let I be the incenter of the triangle ABC. Let the incircle of ABC touch the sides BC, CA, AB at K, L, M respectively. The line through B parallel to MK meets the lines LM and LK at R and S respectively. Prove that the angle RIS is acute.
Solution
We show that RI2 + SI2 - RS2 > 0. The result then follows from the cosine rule.
BI is perpendicular to MK and hence also to RS. So IR2 = BR2 + BI2 and IS2 = BI2 + BS2. Obviously RS = RB + BS, so RS2 = BR2 + BS2 + 2 BR·BS. Hence RI2 + SI2 - RS2 = 2 BI2 - 2 BR·BS. Consider the triangle BRS. The angles at B and M are 90 - B/2 and 90 - A/2, so the angle at R is 90 - C/2. Hence BR/BM = cos A/2/cos C/2 (using the sine rule). Similarly, considering the triangle BKS, BS/BK = cos C/2/cos A/2. So BR·BS = BM·BK = BK2. Hence RI2 + SI2 - RS2 = 2(BI2 - BK2) = 2 IK2 > 0.
(C) John Scholes
jscholes@kalva.demon.co.uk
26 Oct 1998
Last corrected/updated 20 Aug 03