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Bulgarian Winter Tournament

Geometry Difficulty 4.7 AIME Prove it Bulgaria

Points $D$ and $E$ are on sides $BC$ and $AC$ of $\triangle ABC$ . Lines $AD$ and $BE$ intersect at point $S$ . Point $F$ is on side $AB$ and lines $FE$ and $FD$ intersect line $l$ passing through $C$ and parallel to $AB$ at points $P$ and $Q$ . Prove that if $CP = CQ$ , then the points $C$ , $S$ and $F$ lie on the same line.

Solution

From the similarities $\triangle PEC \sim \triangle FEA$ and $\triangle CDQ \sim \triangle BDF$ we obtain that

$\frac{CP}{AF} = \frac{CE}{AE} \quad \text{and} \quad \frac{CQ}{BF} = \frac{CD}{BD}.$
Therefore $\frac{CP}{CQ} = \frac{CE}{AE} \cdot \frac{AF}{BF} \cdot \frac{BD}{DC}$ . By condition $CP = CQ$ , it follows that
$\frac{CE}{AE} \cdot \frac{AF}{BF} \cdot \frac{BD}{DC} = 1.$
Applying Cheva's theorem for $\triangle ABC$ and the points $F$ , $D$ , and $E$ , it follows that the lines $AD$ , $BE$ , and $CF$ intersect in one point, i.e. point $S$ lies on $CF$ . $\square$

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