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Slovenian Mathematical Olympiad, 2008

Geometry Difficulty 5.9 AIME, harder Prove it Slovenia

Let $\triangle ABC$ be an isosceles triangle with the apex at $B$ and choose a point $D$ on the altitude from $B$ such that $AC$ is tangent to the circumcircle $\mathcal{K}$ of the triangle $ABD$ . Let $E$ be a point on $\mathcal{K}$ such that the chord $DE$ is perpendicular to the chord $AB$ . Prove that triangles $ABE$ and $ABC$ are congruent.

Solution

Let $T$ be the intersection of the chords $DE$ and $AB$ . We know that $DTB$ is a right triangle. Let $\angle CAB = \angle ACB = \alpha$ . The line $AC$ is tangent to the circumcircle, so the angle $\angle CAB$ is equal to the corresponding angle $\angle AEB$ over the chord $AB$ . Thus, $\angle AEB = \alpha$ .

We have $\angle ABD = \frac{\pi}{2} - \alpha$ , so $\angle TDB = \alpha$ and $\angle EDB = \alpha$ . This angle is equal to $\angle EAB$ because they are both the inscribed angles over $BE$ . So, $\angle BAE = \alpha = \angle BEA$ and the triangle $ABE$ is isosceles with the apex at $B$ . Triangles $ABE$ and $ABC$ have congruent angles and a common side $AB$ next to the corresponding two angles, so they are congruent.

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