Track / Stage 5 / 82 of 400 #682 of 1964

Problem 682

AIME late

Geometry Difficulty 5.1 Prove it Kanada · Canada · 2011

Let $ABCD$ be a cyclic quadrilateral whose opposite sides are not parallel, $X$ the intersection of $AB$ and $CD$ , and $Y$ the intersection of $AD$ and $BC$ . Let the angle bisector of $\angle AXD$ intersect $AD$ , $BC$ at $E$ , $F$ respectively and let the angle bisector of $\angle AYB$ intersect $AB$ , $CD$ at $G$ , $H$ respectively. Prove that $EGFH$ is a parallelogram.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Since $ABCD$ is cyclic, $\triangle XAC \sim \triangle XDB$ and $\triangle YAC \sim \triangle YBD$ . Therefore,
$\frac{XA}{XD} = \frac{XC}{XB} = \frac{AC}{DB} = \frac{YA}{YB} = \frac{YC}{YD}.$
Let $s$ be this ratio. Therefore, by the angle bisector theorem,
$\frac{AE}{ED} = \frac{XA}{XD} = \frac{XC}{XB} = \frac{CF}{FB} = s,$
and
$\frac{AG}{GB} = \frac{YA}{YB} = \frac{YC}{YD} = \frac{CH}{HD} = s.$
Hence, $\frac{AG}{GB} = \frac{CF}{FB}$ and $\frac{AE}{ED} = \frac{DH}{HC}$ . Therefore, $EH \parallel AC \parallel GF$ and $EG \parallel DB \parallel HF$ . Hence, $EGFH$ is a parallelogram. $\square$

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