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Problem 701

AMC 10/12, early questions

Geometry Difficulty 3.8 Prove it Berkeley Math Circle: Monthly Contest 7 · United States

Given a quadrilateral $ABCD$ , show that the midpoints of its four edges form the vertices of a parallelogram.

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

Solution:

Let $M, N, P, Q$ be the midpoints of $AB, BC, CD$ , and $DA$ , respectively. Then $MN$ is the midline of $\triangle ABC$ opposite $AC$ , so it is parallel to $AC$ and of length $\frac{1}{2} AC$ . Similarly, $PQ$ is the midline of $\triangle ACD$ , so $PQ \parallel AC$ and $PQ = \frac{1}{2} AC$ . Thus, the opposite sides $MN$ and $PQ$ are equal and parallel, and similarly, $NP$ and $QM$ are equal and parallel. Thus, $MNPQ$ is a parallelogram.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.