Track / Stage 3 / 55 of 260 #695 of 2604

Problem 695

AMC 10/12, early questions

Geometry Difficulty 3.2 Prove it Junior Turkish Mathematical Olympiad · Turkey

Let $ABCD$ be a parallelogram. Suppose that a point $P$ is chosen on the arc of the circumcircle of $ABC$ not containing $A$ ; a point $Q$ is chosen on the extension of the segment $AC$ on the side $C$ such that $\angle PBC = \angle CDQ$ . Show that the circumcircle of $APQ$ is tangent to the line $AB$ .

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

The equalities $\angle APB = \angle ACB = \angle QAD$ and $\angle ABP = \angle QDA$ imply the similarity $APB \sim QAD$ . Hence $AP/AQ = PB/AD = BP/BC$ , then the equality $\angle PBC = \angle PAQ$ implies the similarity $BPC \sim APQ$ . Therefore $\angle APQ = \angle BPC = \angle BAQ$ , thus the circle $(APQ)$ is tangent to the line $AB$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.