Track / Stage 6 / 5 of 400 #1005 of 1964

Problem 1005

National olympiad, first round

Geometry Difficulty 6.0 Prove it

Two points $A$ and $C$ are marked on a circle; assume the tangents to the circle at $A$ and $C$ meet at $P$ . Let $B$ be another point on the circle, ans suppose $P B$ meets the circle again at $D$ . Show that $A B \cdot C D=B C \cdot D A$ .

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

(Problem source: Aops) Since $P A$ is tangent to the circle and $\angle A B D$ is the inscribed angle opposite $A D, \angle P A D=\angle A B P$ . Since $\angle A P B$ is shared, it follows from AA similarity that $\triangle P A D \sim \triangle P B A$ . Thus we get

$\frac{A B}{A D}=\frac{P B}{P A}$

Similarly, $\triangle P C D \sim \triangle P B C$ , so

$\frac{B C}{C D}=\frac{P B}{P C}$

Since $P A$ and $P C$ are tangents from the same point to the same circle, they are the same length. Thus,

$\frac{A B}{A D}=\frac{B C}{C D} \Longrightarrow A B \cdot C D=B C \cdot D A$

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