Track / Stage 6 / 181 of 400 #1181 of 1964

Problem 1181

National olympiad, first round

Geometry Difficulty 6.2 Prove it

As shown in the figure, quadrilateral $ABCD$ is inscribed in a circle, the extensions of $AB$ and $DC$ meet at $E$ , the extensions of $AD$ and $BC$ meet at $F$ , $P$ is any point on the circle, $PE$ and $PF$ intersect the circle at $R$ and $S$ respectively. If the diagonals $AC$ and $BD$ intersect at $T$ , prove that: $R, S, T$ are collinear.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Proof: As shown in the figure, connect $PD, AS, RC, BR, AP, SD$ .

From $\triangle EBR \sim \triangle EPA$ and $\triangle FDS \sim \triangle FPA$ , we have
$\frac{BR}{PA} = \frac{EB}{EP}, \quad \frac{PA}{DS} = \frac{FP}{FD}$
Multiplying these two equations, we get
$\frac{BR}{DS} = \frac{EB \cdot FP}{EP \cdot FD} \quad (1)$

From $\triangle ECR \sim \triangle EPD$ and $\triangle FDP \sim \triangle FSA$ , we have
$\frac{CR}{PD} = \frac{EC}{EP}, \quad \frac{PD}{AS} = \frac{FP}{FA}$
Multiplying these two equations, we get
$\frac{CR}{AS} = \frac{EC \cdot FP}{EP \cdot FA} \quad (2)$

From (1) and (2), we have
$\frac{BR \cdot AS}{DS \cdot CR} = \frac{EB \cdot FA}{EC \cdot FD}$
Thus,
$\frac{BR \cdot CD \cdot SA}{RC \cdot DS \cdot AB} = \frac{EB \cdot AF \cdot DC}{BA \cdot FD \cdot CE}$

By Menelaus' theorem, we have
$\frac{EB \cdot AF \cdot DC}{BA \cdot FD \cdot CE} = 1 \quad (4)$
From (3) and (4), we get
$\frac{BR \cdot CD \cdot SA}{RC \cdot DS \cdot AB} = 1$
Therefore, $BD, RS, AC$ intersect at one point, which means $R, T, S$ are collinear.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.