Let the medians of the triangle $A B C$ meet at $M$. Let $D$ and $E$ be different points on the line $B C$ such that $D C=C E=A B$, and let $P$ and $Q$ be points on the segments $B D$ and $B E$, respectively, such that $2 B P=P D$ and $2 B Q=Q E$. Determine $\angle P M Q$.

Answer: $\angle P M Q=90^{\circ}$.
Draw the parallelogram $A B C A^{\prime}$, with $A A^{\prime} \| B C$. Then $M$ lies on $B A^{\prime}$, and $B M=\frac{1}{3} B A^{\prime}$. So $M$ is on the homothetic image (centre $B$, dilation $1 / 3$ ) of the circle with centre $C$ and radius $A B$, which meets $B C$ at $D$ and $E$. The image meets $B C$ at $P$ and $Q$. So $\angle P M Q=90^{\circ}$.

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