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Berkeley Math Circle Monthly Contest 7

Geometry Difficulty 4.7 AIME Prove it United States

Problem:

Let $ABCD$ be a trapezoid with $AB \parallel CD$ . Let $M$ and $N$ be the respective midpoints of $AB$ and $CD$ . Diagonals $AC$ and $BD$ meet at $P$ , and lines $AD$ and $BC$ meet at $Q$ . Prove that $M$ , $N$ , $P$ , and $Q$ are collinear.

Solution

Solution:

Note that $\triangle QAB \sim \triangle QDC$ (by corresponding angles). Since $QM$ and $QN$ are corresponding medians, we have $\angle AQ M = \angle DQ N$ and hence $Q$ , $M$ , and $N$ are collinear.

Similarly, note that $\triangle PAB \sim \triangle PCD$ . Since $PM$ and $PN$ are corresponding medians, we have $\angle AP M = \angle CP N$ and hence $P$ , $M$ , and $N$ are collinear. Since $P$ and $Q$ lie on line $MN$ , all four of these points are collinear.

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