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Irish Mathematical Olympiad, 2014

Geometry Difficulty 5.5 AIME, harder Prove it Ireland

A square $ABCD$ is inscribed in a circle. Let $E$ be the midpoint of $AD$ . The line $CE$ meets the circle again at $F$ . The lines $FB$ and $AC$ meet at $G$ and the line $GE$ meets the arc $AFD$ of the circle at $K$ .
Find, with proof, the measure of $\angle KAD$ .

Solution

Let $O$ be the intersection point of the diagonals $AC$ and $BD$ , which also is the centre of the circle. Because $A$ , $B$ , $C$ , $F$ are on a circle, we have $\angle BFC = \angle BAC = 45^\circ$ . Since $\angle CAD = 45^\circ$ , we obtain $\angle BFC = \angle CAD$ and hence $\angle AEGF$ is cyclic.

This implies that $\angle AEG = \angle AFG = \angle AFC - \angle BFC = 90^\circ - 45^\circ = 45^\circ$ . Because $\angle ADB = 45^\circ$ , too, the lines $EG$ and $DO$ are parallel.
Because $E$ is the midpoint of $AD$ , $G$ is the midpoint of $AO$ . As $DB$ is perpendicular to $AC$ we obtain now that $KG$ is the perpendicular bisector of $AO$ . Hence $|AK| = |OK|$ . As $|OK| = |OA|$ (both are radii), we see now that $AOK$ is an equilateral triangle. We obtain that $\angle OAK = 60^\circ$ and so
$\angle KAD = \angle OAK - \angle CAD = 60^\circ - 45^\circ = 15^\circ.$

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