Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.7 AIME, harder Prove it

Let $A, B, D, E, F, C$ be six points on a circle in that order, satisfying $AB = AC$ . Line $AD$ intersects $BE$ at point $P$ , line $AF$ intersects $CE$ at point $R$ , line $BF$ intersects $CD$ at point $Q$ , line $AD$ intersects $BF$ at point $S$ , and line $AF$ intersects $CD$ at point $T$ . Point $K$ lies on segment $ST$ such that $\angle SKQ = \angle ACE$ . Prove that $\frac{SK}{KT} = \frac{PQ}{QR}$ .
(30th China Mathematical Olympiad)

Solution

Prove as shown in Figure 1, construct $TH \parallel KQ$ , intersecting $BF$ at point $H$ , and connect $RH$ , $CF$ , and $FD$ .
By Pascal's theorem, points $P$ , $Q$ , and $R$ are collinear.
By $AB = AC \Rightarrow \angle AFB = \angle ADC$
$\Rightarrow S$ , $D$ , $F$ , and $T$ are concyclic
$\Rightarrow \angle TSF = \angle CDF = \angle CAF$ .
Also, $AB = AC \Rightarrow \angle TFS = \angle CFA$ .
Therefore, $\triangle FTS \sim \triangle FCA$ .
By $\angle STH = \angle SKQ = \angle ACR$ , then $H$ and $R$ are corresponding points of the similar triangles.
$\begin{array}{l} \text{Hence } \frac{FH}{FS} = \frac{FR}{FA} \Rightarrow RH \parallel AP \\ \Rightarrow \frac{SK}{KT} = \frac{SQ}{QH} = \frac{PQ}{QR}. \end{array}$

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