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Hong Kong competition problems

Geometry Difficulty 4.6 AIME Prove it Hong Kong

Let $ABC$ be a triangle with $AB = AC$ . A circle $\Gamma$ lies outside triangle $ABC$ and is tangent to line $AC$ at $C$ . Point $D$ lies on $\Gamma$ such that the circumcircle of triangle $ABD$ is internally tangent to $\Gamma$ . Segment $AD$ meets $\Gamma$ again at $E$ . Prove that $BE$ is tangent to $\Gamma$ .

Solution

Note that $AB^2 = AC^2 = AE \times AD$ . This implies $\triangle ABE \sim \triangle ADB$ . It follows that $\angle AEB = \angle ABD$ . Let $BD$ meet $\Gamma$ again at $F$ . Consider the homothety with centre $D$ taking $\Gamma$ to the circumcircle of $\triangle ABD$ . Since $EF$ is mapped to $AB$ , the lines $EF$ and $AB$ are parallel. Hence, we have
$\angle AEB = \angle ABD = \angle EFD = \angle ECD.$
This implies $BE$ is tangent to $\Gamma$ .

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