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Geometry Difficulty 8.5 Shortlist Prove it Saudi Arabia

Let $ABC$ be an acute triangle. The line through $A$ perpendicular to $BC$ intersects $BC$ at $D$ . Let $E$ be the midpoint of $AD$ and $\omega$ the circle of center $E$ and radius $AE$ . The line $BE$ intersects $\omega$ at $X$ such that $X$ and $B$ are not on the same side of $AD$ and the line $CE$ intersects $\omega$ at $Y$ such that $C$ and $Y$ are not on the same side of $AD$ . If two intersection points of the circumcircles of triangles $BDX$ and $CDY$ lie on the line $AD$ , prove that $AB = AC$ .

Solution

Note that the condition of the problem is that $AD$ is the radical axis of two circles ( $BDX$ ) and ( $CDY$ ) which implies that $E$ has the same power to these circles. This gives us
$EB \cdot EX = EC \cdot EY.$
However, $EX = EY$ because $E$ is the center of $\omega$ and this means that $BE = CE$ . From this, we conclude that $DE$ is the perpendicular bisector of $BC$ and leads to $AB = AC$ . $\square$

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.