Estonija, 2010

Geometry Difficulty 5.1 AIME, harder Prove it Estonia

In an acute triangle $ABC$ the angle $C$ is greater than the angle $A$ . Let $AE$ be a diameter of the circumcircle of the triangle. Let the intersection point of the ray $AC$ and the tangent of the circumcircle through the vertex $B$ be $K$ . The perpendicular to $AE$ through $K$ intersects the circumcircle of the triangle $BCK$ for the second time at point $D$ . Prove that $CE$ bisects the angle $BCD$ .

Solution

Since $AE$ is a diameter of the circumcircle of the triangle $ABC$ , $\angle ACE = \angle ECK = 90^\circ$ . So it suffices to show that $\angle ACB = \angle DCK$ (Fig. 19).

Let $L$ be the point of intersection of lines $AE$ and $DK$ . Then $\angle BAC = \angle CBK = \angle CDK$ by the inscribed angles theorem. Also $\angle ABC = \angle AEC = \angle CKD$ where the latter equality follows from the similarity of the right triangles $ACE$ and $ALK$ . Hence the two triangles $ABC$ and $DKC$ are similar, and therefore $\angle ACB = \angle DCK$ .

Fig. 19

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