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Saudi Arabian Mathematical Competitions, 2012

Geometry Difficulty 4.5 AIME Prove it Saudi Arabia

The median $AM$ , $M \in BC$ , of the acute-angled triangle $ABC$ intersects the circumcircle of triangle at $D$ . Let $E$ be the symmetric point of $A$ with respect to $M$ . Prove that $BC$ is the common tangent of the circumcircles of triangles $BDE$ and $CDE$ .

Solution

$ABEC$ is a parallelogram. We have $\overline{CBD} \equiv \overline{CAD} \equiv \overline{AEB}$ , so $BC$ is tangent to the circumcircle of triangle $BDE$ . Similarly, from $\overline{BCD} \equiv \overline{BAD} \equiv \overline{AEC}$ , it follows that $BC$ is tangent to the circumcircle of triangle $CDE$ .

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