IMO 1977 LL USA55

Origin: USA

Problem

Through a point O on the diagonal BD of a parallelogram ABCD, segments MN parallel to AB, and PQ parallel to AD, are drawn, with M on AD, and Q on AB. Prove that diagonals AO, BP, DN (ex- tended if necessary) will be concurrent.

Solution

The statement is true without the assumption that O \inBD. Let BP \cap DN = {K}. If we denote −−\to AB = a, −−\to AD = b and −\to AO = \alphaa + \betab for some \alpha, \beta \inR, 1/\alpha + 1/\beta ̸= 1, by straightforward calculation we obtain that

−−\to AK = \alpha \alpha + \beta −\alpha\beta a + \beta \alpha + \beta −\alpha\beta b = \alpha + \beta −\alpha\beta −\to AO. Hence A, K, O are collinear.