A Solution to Crux Mathematicorum Problem 5088

The original problem statement is intentionally not reproduced here. This article follows its notation and presents only the independently prepared solution published here and its diagram. The complete statement is available in the official issue of Crux Mathematicorum.

Solution

Parallelogram ABCD with BC equal to BD and M the midpoint of CD. E, F, and G are the perpendicular feet from M to the lines BD, AD, and BC. The circle centered at M passes through E, F, G, and K; AK is tangent at K, and N is the intersection of EG and BM. — Figure 1. The circle centered at M and the auxiliary points used in the proof.Select the diagram to enlarge it.

Proof

Let $M$ be the midpoint of $CD$ . Let $F$ and $G$ be the feet of the perpendiculars from $M$ to $AD$ and $BC$ , respectively. Let $N=EG\cap BM$ .

Since $MD=MC$ and $\measuredangle MDF=\measuredangle MCB=\measuredangle EDM$ , the three right triangles $MDF$ , $MCG$ , and $MDE$ are congruent. Hence $EM=FM=GM$ , and $E$ , $F$ , $G$ , and $K$ lie on the circle $\omega$ , centered at $M$ .

$A$ , $K$ , $M$ , and $F$ are concyclic, since $\measuredangle AKM=\measuredangle AFM=\frac{\pi}{2}$ . Since $BC=BD$ and $M$ is the midpoint of $CD$ , we have $BM\perp CD$ , and hence $BM\perp AB$ . Therefore, $A$ , $B$ , $M$ , and $F$ are concyclic, since $\measuredangle ABM=\measuredangle AFM=\frac{\pi}{2}$ . Thus $A$ , $B$ , $F$ , $K$ , and $M$ are concyclic.

Moreover, $BG$ and $BE$ are tangents to $\omega$ at $G$ and $E$ , respectively. It follows that $BN\perp EG$ .

Since $F$ , $M$ , and $G$ are collinear,

\begin{aligned} \measuredangle NEK &=\measuredangle GEK\\ &=\measuredangle GFK\\ &=\measuredangle MFK\\ &=\measuredangle MBK\\ &=\measuredangle NBK. \end{aligned}

Thus $N$ , $K$ , $E$ , and $B$ are concyclic. Therefore, $\measuredangle BKE=\measuredangle BNE=\frac{\pi}{2}$ . This completes the proof.

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