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Bulgarian Winter Tournament

Geometry Difficulty 5.9 AIME, harder Prove it Bulgaria

The inscribed circle in $\triangle ABC$ ( $AC \neq BC$ ) is tangent to its sides $AB$ , $BC$ , and $CA$ at points $D$ , $E$ , and $F$ , respectively. Let $P$ be the foot of the perpendicular from $D$ to $EF$ ( $P \in EF$ ). If the circles circumscribed about $\triangle ABC$ and $\triangle EFC$ intersect for the second time at point $Q$ , prove that $\angle PQC = 90^\circ$ .

(Stoyan Boev)

Solution

From the fact that $C$ lies on the circle circumscribed about $\triangle FEQ$ and $CE = CF$ it follows that $CQ$ is an exterior bisector of $\angle FQE$ and it remains to prove that $QP$ is a bisector of $\angle FQE$ , i.e. $QF : QE = FP : PE$ . From $\angle QFC = \angle QEC$ and $\angle QAC = \angle QBC$ it follows that $\triangle QFA \sim \triangle QEB$ , i.e.
$QF : QE = AF : BE = AD : BD.$
Let $I$ be the center of the circle $k$ inscribed in $\triangle ABC$ and the line $DP$ intersects $k$ for the second time at point $R$ . Then
$\angle REF = \angle RDF = 90^\circ - \angle DFE = 90^\circ - \angle DIB = \angle IBA$
and analogously $\angle RFE = \angle IAB$ , i.e. $\triangle FER \sim \triangle ABI$ . But $RP \perp EF$ , $ID \perp AB$ , i.e. $P$ and $D$ are corresponding elements in similar triangles and $FP : PE = AD : BD$ , which completes the proof. $\square$

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