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, 2023

Geometry Difficulty 5.0 AIME, harder Prove it Turkey

Let $ABC$ be a triangle and $D$ , $E$ be points on segments $AB$ , $AC$ respectively, such that $DE \parallel BC$ . Let the circumcircle of $ABC$ meet the circumcircles of $BDE$ and $CDE$ again at $K$ , $L$ respectively. Let $T$ be the intersection of the lines $BK$ and $CL$ . Prove that $TA$ is tangent to the circumcircle of $ABC$ .

Solution

The radical axes of the circles $(ABC)$ , $(BDE)$ , $(CDE)$ must be concurrent at $T$ ; hence $T$ , $D$ , $E$ are collinear. Moreover, $TD \cdot TE = TB \cdot TK$ ; hence the power of $T$ with respect to the circles $(ABC)$ , $(ADE)$ are equal and it lies on their radical axis. Since $DE$ and $BC$ are parallel, the radical axis of $(ABC)$ , $(ADE)$ is the common tangent at $A$ ; hence $TA$ is tangent to $(ABC)$ .

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