A Solution to Crux Mathematicorum Problem 5088

Problem 5088

Proposed by To An Ky

Let $ABCD$ be a parallelogram satisfying $BC=BD$ . Let $\omega$ be a circle centered at the midpoint of segment $CD$ , touching side $BD$ at $E$ . From $A$ , construct segment $AK$ tangent to $\omega$ at $K$ $(K\notin\overleftrightarrow{AD})$ . Show that $\angle BKE=90^\circ$ .

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.

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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Solution

Diagrams and complete source