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Saudi Arabian Mathematical Competitions, 2012

Geometry Difficulty 4.4 AIME Prove it Saudi Arabia

Consider the isosceles triangle $ABC$ with $AB = AC$ . A semicircle of diameter $EF$ situated on the side $BC$ , is tangent to the sides $AB$ and $AC$ at $M$ and $N$ , respectively. The line $AE$ intersects the semicircle at $P$ . Prove that the line $PF$ passes through the midpoint of the chord $MN$ .

Solution

Let $O$ be the center of the semicircle and let $R$ be the midpoint of segment $MN$ .

In triangle $ANO$ we have $AN^2 = AR \cdot AO$ . Using the power of the point $A$ with respect to the circle we get
$AM^2 = AP \cdot AE = AN^2 = AR \cdot AO. \quad (1)$
From (1) it follows that the quadrilateral $PROE$ is cyclic, hence $RP \perp AE$ . Since $FP \perp AE$ , we get that the points $F$ , $R$ , $P$ are collinear.

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