# Mid Euro 2009 Problem 9

Problem 9 of the Middle European Mathematical Olympiad 2009

Let ABCD be a parallelogram with ∠BAD = 60° and denote by E the intersection of its diagonals. The circumcircle of triangle ACD meets the line BA at K≠A, the line BD at P≠D and the line BC at L≠C. The line EP intersects the circumcircle of triangle CEL at points E and M. Prove that triangles KLM and CAP are congruent.

Solution by Steve Dinh, a.k.a. Vo Duc Dien Since K, L, A, C and M, L, E, C are concyclic, we have KL/AC = BL/AB, AP/CD = AE/DE and ML/EC = BL/BE or ML/AP = ECxBLxDE \over BExAExCD

but since E is the intersection of the diagonals of the parallelogram ABCD, BE = DE, AE = EC.

It follows that ML/AP = BL/CD = BL/AB = KL/AC (*)

We also have ∠LPD = ∠LCD = ∠BAD = 60°

Now chase the angle ∠KLM = ∠KLP + ∠MLP = ∠KLP + 60° - ∠LMP = ∠KLP + 60° - ∠LME = ∠KLP + 60° - ∠LCE = ∠KLP + ∠ACD = ∠KLP + ∠KLC (since AB || CD and KC = AD) = ∠CLP = ∠CAP (subtending arc CP).

Combined with (*), the two triangles KLM and CAP are similar.

Furthermore, since DL = AC (diagonals of isosceles trapezoid ADCL) = DK (diagonals of isosceles trapezoid ADCK). But since ∠BAD = 60° subtends arc DK, it follows that KDL is an equilateral triangle, and KL = DK = AC. This makes the two already similar triangles KLM and CAP congruent.