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11-th Czech-Slovak-Polish Match, 2011

Geometry Difficulty 5.1 AIME, harder Prove it Czech-Polish-Slovak Mathematical Match

In a convex quadrilateral $ABCD$ let $M$ and $N$ be midpoints of sides $AD$ and $BC$ , respectively. Points $K$ and $L$ are chosen on sides $AB$ and $CD$ , respectively, in such a manner that $\angle MKA = \angle NLC$ . Prove that if lines $BD$ , $KM$ and $LN$ meet at one point, then
$\angle KMN = \angle BDC, \quad \angle LNM = \angle ABD$

Solution

Let $P$ be the midpoint of $BD$ and $Q$ be the common point of lines $BD$ , $KM$ and $LN$ . Without losing generality assume that point $B$ lies between $Q$ and $D$ . By Tales theorem, $PM \parallel AB$ and $PN \parallel CD$ . Therefore, $\angle PNL = \angle NLC = \angle MKA = \angle KMP$ . Note that this implies points $Q, M, P, N$ to be concyclic, as $\angle QNP + \angle QMP = 180^{\circ}$ and points $M, N$ lie on different sides of the line $BD$ . Therefore, $\angle KMN = \angle QMN = \angle QPN = \angle BDC$ . Moreover, $\angle LNM = 180^{\circ} - \angle QNM = 180^{\circ} - \angle QPM = \angle MPD = \angle ABD$ . $\square$

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