On the sides $AC$ and $AB$ of an acute $\triangle ABC$ are chosen points $M$ and $N$ respectively. Point $P$ is an intersection point of the segments $BM$ and $CN$ and point $Q$ is an inner point for the quadrilateral $ANPM$, for which $\angle BQC = 90^\circ$ and $\angle BQP = \angle BMQ$. If the quadrilateral $ANPM$ is inscribed in a circle, prove that $\angle QNC = \angle PQC$.
Problem
Source: IX International Festival of Young Mathematicians Sozopol, Theme for 10-12 grade
Tags: geometry
Ari004
24.09.2018 13:17
The given angle condition implies $BQ$ is tangent to $(PQM)$,so $BQ^2=BP \times BM$,since $ANPM$ is cyclic,it implies $BQ^2=BP \times BM=BN \times BA$,so $BQ$ is also tangent to $(ANQ)$.Now,we use a lemma: Consider $\bigtriangleup ABC$ and $\bigtriangleup CDE$,then $(ABC)$ and $(CDE)$ are tangent to each other if and only if $\angle BAC+\angle CDE=\angle BCE$. Proof of if direction:Let $CF$ be the common tangent to $(ABC)$ and $CDE$.Then $\angle BAC+\angle CDE=\angle BCF+\angle ECF=\angle BCE$,which proves our claim. Proof of only if direction:Let $CF$ be the tangent to $(ABC)$.Since,$\angle BAC=\angle BCF$,$\implies \angle ECF=\angle CDE$,hence $CF$ is the common tangent of $(ABC)$ and $(CDE)$. Coming back to our problem,notice that $\angle AQN=360^{\circ}-(\angle NQP+\angle PQM+\angle AQM)$.Using our lemma for $\bigtriangleup AQN$ and $\bigtriangleup PQM$,we have $\angle NQP=\angle NAQ+\angle QMP$ and $\angle AQM=\angle ANQ+\angle QPM$.Hence, $\angle AQN=360^{\circ}-(\angle NAQ+\angle QMP+\angle ANQ+\angle QPM+\angle PQM) =360^{\circ}-(180^{\circ}-\angle QNP-\angle AMQ+180^{\circ}-\angle MAQ-\angle NPQ+\angle PQM)=\angle QNP+\angle MAQ-\angle PQM+\angle AMQ+\angle NPQ=\angle AMQ+\angle NPQ$. So,$(PQN)$ and $(AQM)$ are tangent to each other.Notice that $C$ lies on the radical axis of $(PQN)$ and $(AQM)$,so $CQ$ must be tangent to both the circles.Hence,$\angle QNC=\angle PQC \blacksquare$.