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Problem 801

AMC 12 late, AIME early

Geometry Difficulty 4.4 Prove it Olimpiada Matemática Rioplatense · Argentina

Let $ABCD$ be a parallelogram. Construct a square $BDXY$ with no interior points in common with the triangle $ABD$ and a square $ACZW$ with no interior points in common with the triangle $ADC$ .
Let $P$ and $Q$ be the centers of the squares $BDXY$ and $ACZW$ respectively. Prove that $AP = DQ$ .

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

Let $N$ be the intersection point of the diagonals of $ABCD$ . As $ABCD$ is a parallelogram, we have that $N$ is the midpoint of $AC$ and the midpoint of $BD$ .

Since $P$ is the center of the square $BDXY$ and $N$ is the midpoint of its side $BD$ , then $DN = NP$ (both equal to a half of $BD$ ) and $D\hat{N}P = 90^\circ$ . Similarly, $AN = NQ$ and $A\hat{N}Q = 90^\circ$ . Then, $A\hat{N}P = 90^\circ + A\hat{N}D = Q\hat{N}D$ . We conclude that the triangles $ANP$ and $QND$ are congruent (by side-angle-side criterion) and, therefore, $AP = DQ$ .

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