Question 63
Let $ABC$ be a triangle such that, if $H$ is the foot of the altitude from $C$, then $AH=3\cdot HB$. Let also:
- $M$ be the midpoint of $AB$;
- $N$ be the midpoint of $AC$;
- $P$ be the point on the opposite side of line $AC$ from $B$ such that $NP=NC$ and $PC=CB$.
Prove that $\angle APM=\angle PBA$.
Supplementary illustration (one configuration):
- $M$ be the midpoint of $AB$;
- $N$ be the midpoint of $AC$;
- $P$ be the point on the opposite side of line $AC$ from $B$ such that $NP=NC$ and $PC=CB$.
Prove that $\angle APM=\angle PBA$.
Supplementary illustration (one configuration):