Track / Stage 4 / 121 of 340 #381 of 1964

Problem 381

AMC 12 late, AIME early

Geometry Difficulty 4.7 Prove it Singapore Mathematical Olympiad (SMO) · Singapore

In a convex quadrilateral $ABCD$ , the diagonals intersect at $O$ , $M$ and $N$ are points on the segments $OA$ and $OD$ respectively. Suppose $MN$ is parallel to $AD$ and $NC$ is parallel to $AB$ . Prove that $\angle ABM = \angle NCD$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

We use the notation $[PQR]$ to denote the area of $\triangle PQR$ .
$MN \parallel AD \Rightarrow [AMN] = [DMN] \Rightarrow [AON] = [DOM].$
$NC \parallel AB \Rightarrow [ACN] = [BCN] \Rightarrow [AON] = [BOC].$
$\text{Thus } [DOM] = [BOC] \Rightarrow [MCD] = [BCD] \Rightarrow MB \parallel DC.$
$\text{Finally } AB \parallel NC, MB \parallel DC \Rightarrow \angle ABM = \angle NCD.$

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