Post Number:#1 by Kazi_Zareer » Sat Feb 04, 2017 12:43 pm
et $ABC$ be a triangle inscribed in circle $(O)$, incenter $I$. Circle $(K)$ touches $CA,AB$ at $E,F$ and touches $(O)$ internally. $AI$ cuts $(O)$ again at $P$. $PQ$ is diameter of $(O)$. $QI$ cuts $BC$ at $D$. $M,N$ are midpoints $DI$ and $KA$. $R$ is on perpendicular bisector of $AQ$ such that $MR\parallel DN$. $J$ is midpoing of $OI$. Prove that $\angle APR=\angle QPJ$.
We cannot solve our problems with the same thinking we used when we create them.