Problem 1145

Solution. Denote by $s$ the line $A D$. Let $T$ be the second intersection point of the circumcircles of $\triangle B D X$ and $\triangle C D Y$. Then $T$ is on the line $s$. Note that $C D Y T$ and $B D X T$ are cyclic. Using this and the fact that $A D$ is perpendicular to $B C$ we obtain:

$$\angle T Y E=\angle T Y C=\angle T D C=90^{\circ}$$

This means that $E Y$ is perpendicular to $T Y$, so $T Y$ must be tangent to $\omega$. We similarly show that $T X$ is tangent to $\omega$. Thus, $T X$ and $T Y$ are tangents from $T$ to $\omega$ which implies that $s$ is the perpendicular bisector of the segment $X Y$. Now denote by $\sigma$ the reflection of the plane with respect to $s$. Then the points $X$ and $Y$ are symmetric with respect to $s$, so $\sigma(X)=Y$. Also note that $\sigma(E)=E$, because $E$ is on $s$. Using the fact that $B C$ is perpendicular to $s$, we see that $B C$ is the reflection image of itself with respect to $s$. Now note that $B$ is the intersection point of the lines $E X$ and $B C$. This means that the image of $B$ is the intersection point of the lines $\sigma(E X)=E Y$ and $\sigma(B C)=B C$, which is $C$. From here we see that $\sigma(B)=C$, so $s$ is the perpendicular bisector of $B C$, which is what we needed to prove.

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