IMO 1987 SL 21

Origin: USS

Problem

The prolongation of the bisector AL (L \inBC) in the acute- angled triangle ABC intersects the circumscribed circle at point N. From point L to the sides AB and AC are drawn the perpendiculars LK and LM respectively. Prove that the area of the triangle ABC is equal to the area of the quadrilateral AKNM.

Solution

Let P be the second point of inter- section of segment BC and the cir- cle circumscribed about quadrilat- eral AKLM. Denote by E the in- tersection point of the lines KN and BC and by F the intersection point of the lines MN and BC. Then \angleBCN = \angleBAN and \angleMAL = \angleMPL, as angles on the same arc. Since AL is a bisector, \angleBCN = \angleBAL = \angleMAL = \angleMPL, and consequently PM \parallelNC. Similarly A B C N E F K M P we prove KP \parallelBN. Then the quadrilaterals BKPN and NPMC are trapezoids; hence SBKE = SNPE and SNPF = SCMF . Therefore SABC = SAKNM.