In the triangle $ABC$ the segment $AD$ is altitude. $M$ is the midpoint of $BC$ and $N$ is the reflection of $M$ in $D$. The circumcircle of $ANM$ meets $AB$ at $P$ and $AC$ at $Q$. Show that $AN, BQ, CP$ are concurrent.
All I have tried to do is angle chasing and hopefully get the sum of the three angles at the point to equal 180 degrees, but I'm stuck. I have also got that AMN is isosceles with AM=AN.
Best Answer
By Ceva's theorem $AN,BQ,CP$ are concurrent iff $$ BN\cdot CQ\cdot AP = CN\cdot AQ\cdot BP\tag{1}$$ and the involved lengths are not difficult to compute. From $BD=c\cos B=\frac{a^2+c^2-b^2}{2a}$ and $BM=\frac{a}{2}$ we get $DM=\frac{b^2-c^2}{2a}$ and $NM=\frac{b^2-c^2}{a}$, from which $CN=\frac{a^2+2b^2-2c^2}{2a}$. Since $CM\cdot CN = CQ\cdot CA$ it follows that $CQ=\frac{a^2+2b^2-2c^2}{4b}$ and $AQ=\frac{-a^2+2b^2+2c^2}{4b}$. In a similar way we may compute $AP$ and $BP$ and check that $(1)$ holds.