An isosceles triangle $ABC$ with $AC = BC$ is given. Let $M$ be the midpoint of the side $AB$ and let $P$ be a point inside the triangle such that $\angle PAB = \angle PBC$. Prove that $\angle APM + \angle BPC = 180 \textdegree $
Problem
Source: 6th Grand Duchy of Lithuania contest
Tags: geometry, circumcircle, geometry proposed
AkshajK
31.12.2014 18:44
Rephrasing so I can copy my solution from elsewhere wrote: Let $\triangle ABC$ be isosceles with base $BC$. Let $P$ be in $ABC$ with $\angle CBP = \angle ACP$. Let $M$ be the midpoint of $BC$. Show that $\angle BPM + \angle CPA = 180$
It is clear $\angle CBP = \angle ACP \implies \angle ABP = \angle BCP$, meaning the circumcircle of $BPC$ is tangent to $AB, AC$ at $B, C$. So from a well-known lemma, $AP$ is the P-symmedian of $\triangle BPC$. So denoting $X = AP \cap BC$ we get $\angle BPX = \angle MPC$ or $\angle BPM = \angle XPC = 180 - \angle APC$ so we are done.
ARCH999
31.12.2014 18:48
My solution: Since $\angle CAB= \angle CBA$, we get $\angle PAC= \angle PBA$. So, $CA,CB$ are tangents to circle $APB$ by equal angles in alternate segments. Thus, if $AP$ is produced to meet $BC$ at $Q$, $PQ$ is the $symmedian$ and $PM$ is the median of triangle $PAB$ through $P$. Since they are isogonal conjugates, $\angle APM= \angle QPB$. So, $\angle APM + \angle BPC = \angle QPB + \angle BPC = 180^\circ$.
ARCH999
31.12.2014 18:49
Sorry, I missed your post.......@Akshajk as I did not use the full editor...