Proving three lines are concurrent 
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. 

 A: 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.
A: One can also solve it without trigonometry: 
First, show that $AP \cdot AB = QA \cdot CA$. This can be done by producing $AD$ until it intersects the circle at $L$ for the second time. Then $\angle \, LPA = \angle \, BDA = 90^{\circ}$ so triangles $LPA$ and $BDA$ are similar and this latter fact implies the ratio $\frac{AB}{AL} = \frac{AD}{AP}$ which is equivalent to the identity $AP \cdot AB = AD \cdot AL\,$. Analogously,  $AD \cdot AL = QA \cdot CA \,$
Second, by the intersecting secant theorem, 
$$BN \cdot BM = PB \cdot AB \,\,\, \text{ and } \,\,\, NC \cdot MC = CA \cdot CQ$$  which means that
$$BN  = \frac{PB \cdot AB}{BM} \,\,\, \text{ and } \,\,\, NC = \frac{CA \cdot CQ}{MC} = \frac{CA \cdot CQ}{BM}$$ because $BM = MC$.
Third, calculate the ratio
$$\frac{BN}{NC}\cdot\frac{CQ}{QA}\cdot\frac{AP}{PB} = \frac{PB \cdot AB}{BM}\cdot \frac{BM}{CA \cdot CQ} \cdot \frac{CQ}{QA}\cdot\frac{AP}{PB} = \frac{PB \cdot AB}{CA \cdot CQ} \cdot \frac{CQ}{QA}\cdot\frac{AP}{PB} = \frac{AP \cdot AB}{QA \cdot CA} $$ However, by $AP \cdot AB = QA \cdot CA$
$$\frac{BN}{NC}\cdot\frac{CQ}{QA}\cdot\frac{AP}{PB} = \frac{AP \cdot AB}{QA \cdot CA}  = 1$$ which, by Ceva's theorem, is possible if and only if the lines $AN, \, BQ$ and $CP$ are concurrent.
