Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 5.1 AIME, harder Prove it Brazil

$ABCD$ is a rhombus. Take points $E, F, G, H$ on sides $AB, BC, CD, DA$ respectively so that $EF$ and $GH$ are tangent to the incircle of $ABCD$ . Show that $EH$ and $FG$ are parallel.

Solution

Let $O$ be the center of the incircle. We show first that $AE \cdot CF = AO^2$ . Let $\angle AOE = \theta$ . Then if $OX$ is the perpendicular from $O$ to $AB$ , we have $\angle AOX = B/2$ and hence $\angle XOE = \theta - B/2$ . If $EF$ touches the circle at $Y$ , $\angle EOY = \angle XOE$ , so $\angle BEF = 2(\theta - B/2) = 2\theta - B$ . Hence $\angle BFE = 180^\circ - 2\theta$ . Hence $\angle CFE = 2\theta$ . So $\angle CFO = \theta$ . So we have established that $\angle AOE = \angle CFO$ . But $\angle EAO = \angle OCF$ (since $ABCD$ is a rhombus), so $AOE$ and $CFO$ are similar. Hence $AE/AO = CO/CF \Leftrightarrow AE \cdot CF = AO^2$ .

Similarly, $AH \cdot CG = AO^2$ . Hence $AH/AE = CF/CG$ . So $AHE$ and $CFG$ are similar. So $\angle AEH = \angle CGF$ , so $EH$ and $FG$ are parallel.

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