Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Prove it

$A B C D$ is a convex quadrilateral in which $A B$ is the longest side. Points $M$ and $N$ are located on sides $A B$ and $B C$ respectively, so that each of the segments $A N$ and $C M$ divides the quadrilateral into two parts of equal area. Prove that the segment $M N$ bisects the diagonal $B D$ .

Solution

Since $[M A D C]=\frac{1}{2}[A B C D]=[N A D C]$ , it follows that $[A N C]=[A M C]$ , so that $M N \| A C$ . Let $m$ be a line through $D$ parallel to $A C$ and $M N$ and let $B A$ produced meet $m$ at $P$ and $B C$ produced meet $m$ at $Q$ . Then

$[M P C]=[M A C]+[C A P]=[M A C]+[C A D]=[M A D C]=[B M C]$

whence $B M=M P$ . Similarly $B N=N Q$ , so that $M N$ is a midline of triangle $B P Q$ and must bisect $B D$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.