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Austrian Mathematical Olympiad

Geometry Difficulty 5.1 AIME, harder Prove it Austria

Let $ABC$ be an isosceles triangle with $AC = BC$ and circumcircle $k$ . The point $D$ lies on the shorter arc of $k$ over the chord $BC$ and is different from $B$ and $C$ . Let $E$ denote the intersection of $CD$ and $AB$ .
Prove that the line through $B$ and $C$ is a tangent of the circumcircle of the triangle $BDE$ .

Solution

We denote the center of the circumcircle of the triangle $BDE$ by $M$ and $\angle BAC = \angle CBA$ by $\alpha$ . Since the quadrilateral $ABDC$ is cyclic, we obtain $\angle BDE = \alpha$ . By the inscribed angle theorem, $\angle BME = 2\alpha$ and thus $\angle EBM = \angle MEB = 90^\circ - \alpha$ . Therefore,
$180^\circ = \angle CBA + \angle MBC + \angle EBM = \alpha + \angle MBC + 90^\circ - \alpha = \angle MBC + 90^\circ$
holds and we get
$\angle MBC = 90^\circ,$
completing the proof.

Figure 1: Problem 6

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