, 2017

Geometry Difficulty 8.5 Shortlist Prove it Hong Kong

Let the internal angle bisector of $\angle BAC$ of $\triangle ABC$ meet side $BC$ at $D$ . Let $\Gamma$ be the circle through $A$ tangent to $BC$ at $D$ . Suppose $\Gamma$ meets sides $AB$ and $AC$ at $E$ and $F$ again, respectively. Lines $BF$ and $CE$ meet $\Gamma$ again at $P$ and $Q$ , respectively. Let $AP$ and $AQ$ intersect side $BC$ at $X$ and $Y$ , respectively. Prove that $XY = \frac{1}{2}BC$ .

Solution

Since $\angle AFD = \angle ADB$ and $\angle DAF = \angle BAD$ , we have $\angle ADF = \angle ABD$ . This implies $\angle AEF = \angle ADF = \angle ABD$ so that $EF \parallel BC$ . It follows that $\angle XBP = \angle EFP = \angle BAX$ . Thus, $XB$ is a tangent to $(ABP)$ . Hence, we have $XB^2 = XP \cdot XA = XD^2$ . This gives $XB = XD$ . Similarly, we have $YC = YD$ . Therefore, $XY = \frac{1}{2}BC$ . $\square$

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