<s>[INST] 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. 

 [/INST] 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.
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