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Geometry Difficulty 8.5 Shortlist Prove it Turkey

$D$ , $E$ , $F$ are points on the sides $AB$ , $BC$ , $CA$ , respectively, of a triangle $ABC$ such that $AD = AF$ , $BD = BE$ and $DE = DF$ . Let $I$ be the incircle of the triangle $ABC$ , and let $K$ be the point of intersection of the line $BI$ and the tangent line through $A$ to the circumcircle of the triangle $ABI$ . Show that $AK = EK$ if $AK = AD$ .

Solution

From $AD = AF$ , $BD = BE$ and $DE = DF$ we obtain $AD \sin(\angle A/2) = BD \sin(\angle B/2)$ . Using this and applying the law of sines to the triangle $AIB$ , we get $AI/BI = AD/BD = AK/BE$ . Since $AK$ is tangent to the circumcircle of $AIB$ , we also have $\angle KAI = \angle ABI = \angle EBI$ . Hence the triangles $KAI$ and $EBI$ are similar. It follows that the points $A, I, E$ are collinear and the points $A, K, E, B$ are concyclic. In particular, $\angle KEA = \angle KBA = \angle EBI = \angle KAE$ and $AK = EK$ .

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