IMO 1982 SL 9

Problem

B3 (GBR 1) Let ABC be a triangle, and let P be a point inside it such that ∡PAC = ∡PBC. The perpendiculars from P to BC and CA meet these lines at L and M, respectively, and D is the midpoint of AB. Prove that DL = DM.

Solution

Let X and Y be the midpoints of the segments AP and BP. Then DY PX is a parallelogram. Since X and Y are the circumcenters of \triangleAPM and \triangleBPL, it follows that XM = XP = DY and Y L = Y P = DX. Furthermore, \angleDXM = \angleDXP + \anglePXM = \angleDXP + 2\anglePAM = \angleDY P + 2\anglePBL

\angleDY P + \anglePY L = \angleDY L. Therefore, the triangles DXM and LY D are con- gruent, implying DM = DL. A B C P M L X Y D