Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.7 National olympiad Prove it Argentina

Let $ABC$ be an acute-angled triangle with $AC > AB$ . Let $\Gamma$ be the circumference circumscribed about the triangle $ABC$ and $D$ the midpoint of the smaller arc $BC$ of $\Gamma$ . Let $E$ and $F$ be points in the segments $AB$ and $AC$ respectively such that $AE = AF$ . Let $P \ne A$ be the second intersection point of the circumference circumscribed about the triangle $AEF$ with $\Gamma$ . Let $G$ and $H$ be the points, different from $P$ , where the lines $PE$ and $PF$ intersect $\Gamma$ , respectively. Let $J$ and $K$ be the intersections of the lines $DG$ and $DH$ with the lines $AB$ and $AC$ respectively. Prove that the line $JK$ passes through the midpoint of $BC$ .

Solution

Let $M$ be the midpoint of the segment $BC$ .

Let $\alpha = \angle AEF$ . Since $AE = AF$ , we have that $\angle AFE = \angle AEF = \alpha$ . Considering the cyclic quadrilaterals $APEF$ and $APDH$ , we have that
$\alpha = \angle AEF = \angle APF = \angle APH = \angle ADH.$
Now, since $\angle AFI = \angle ADK$ , then the quadrilateral $IFKD$ is cyclic. Hence, $\angle FKD + \angle FID = 180^\circ$ . Also, $\angle EAI = \angle FAI$ , because $D$ is the midpoint of the arc $BC$ ; then, $\angle AIF = 90^\circ$ . Therefore, $\angle FKD = 90^\circ$ and, then, $DK$ is perpendicular to $AC$ .

On the other hand, the quadrilaterals $APEF$ and $APGD$ are cyclic and
$\angle ADG = 180^\circ - \angle APG = 180^\circ - \angle APE = \angle AFE = \alpha.$
Since $\angle AEI = \angle ADJ = \alpha$ , we have that the quadrilateral $DJEI$ is cyclic and $\angle AJD = 180^\circ - \angle EID = 90^\circ$ . Then, $DJ$ is perpendicular to $AB$ .

Finally, since $D$ is the midpoint of the arc $BC$ and $M$ is the midpoint of the corresponding chord, $DM$ is perpendicular to $BC$ .

Using the Simson line of the point $D$ , we conclude that $J$ , $M$ and $K$ are collinear.

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