Problem

Source: CMO 2024 P5

Tags: geometry, circumcircle



In acute $\triangle {ABC}$, ${K}$ is on the extention of segment $BC$. $P, Q$ are two points such that $KP \parallel AB, BK=BP$ and $KQ\parallel AC, CK=CQ$. The circumcircle of $\triangle KPQ$ intersects $AK$ again at ${T}$. Prove that: (1) $\angle BTC+\angle APB=\angle CQA$. (2) $AP \cdot BT \cdot CQ=AQ \cdot CT \cdot BP$. Proposed by Yijie He and Yijuan Yao