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Taiwan IMO Selection Camp

Geometry Difficulty 4.5 AIME Prove it Taiwan

On the extension of side $BC$ of $\triangle ABC$ , take a point $D$ such that $CD = AC$ . The circumcircle of $\triangle ACD$ intersects the circle with $BC$ as diameter at points $C$ , $P$ , the line $BP$ intersects $AC$ at $E$ , and the line $CP$ intersects $AB$ at $F$ . Prove that the three points $D$ , $E$ , $F$ are collinear.

Solution

As shown in the figure. Since point $P$ lies on the circumcircle of $\triangle ACD$ , and $AC = CD$ , we have
$\angle APF = \angle ADC = \angle CAD = \angle CPD,$
that is, $PC$ is the external bisector of $\angle APD$ . Thus, letting the line $AP$ meet $BC$ at $Q$ , $PC$ bisects $\angle QPD$ . Also, since point $P$ lies on the circle with $BC$ as diameter, $BP \perp PC$ , so $BP$ is the external bisector of $\angle QPD$ , from which we obtain
$\frac{BD}{DC} = -\frac{BQ}{QC}.$
On the other hand, since $AQ$ , $BE$ , $CF$ meet at point $P$ , by Ceva's theorem we have
$\frac{BQ}{QC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1,$
hence
$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = -\frac{BQ}{QC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = -1$
and then by Menelaus's theorem, the three points $D$ , $E$ , $F$ are collinear.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.