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Geometry Difficulty 8.3 Shortlist Prove it Turkey

Let $ABC$ be a triangle satisfying $|AB| = |AC|$ . Let $D$ be a point on smaller arc $AC$ of circumcircle $\omega$ of $ABC$ . Let $E$ be the symmetric point of $B$ with respect to the line $AD$ . The line $BE$ intersects $\omega$ at $F$ . The tangent line of $\omega$ at $F$ intersects the line $AC$ at $K$ and the lines $DF$ and $AB$ intersect at $L$ . Show that the points $K, L, E$ are collinear.

Solution

Let $M$ be a point on $AD$ such that $D \in [AM]$ . Since $\angle EDA = \angle ADB = \angle ACB = \angle ABC = \angle CDM$ , we conclude that the points $C$ , $D$ , $E$ are collinear.

Using Pascal theorem for the points $D$ , $C$ , $A$ , $B$ , $F$ , $F$ , we find that the points $E = DC \cap BF$ , $K = CA \cap FF$ , $L = AB \cap FD$ are collinear, and we are done. (Here $FF$ denotes the tangent line of $\omega$ at $F$ .)

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.