If $AC=BC$ and $\angle {PAB}=\angle {PBC}$. Prove that $\angle{APM}+\angle{BPC}$=$180°$

Question

In $\Delta ABC$, $AC=BC$. $P$ is a point inside $\Delta ABC$ such that $\angle {PAB}=\angle {PBC}$. $M$ is the midpoint of $AB$. We need to show $\angle {APM}=\angle {BPC}$.

I am very puzzle while solving this.How to solve this?I failed to find any thing. My problem is to do some good thing about the $'M'$. This is the weird part of this problem.I want to do this will classical geometry(i.e.by some synthetic techniques,not by bash like coordinate,complex number,barycentric coordinates ).Please help me

Answer 1

Let $CP$ and $AB$ meet at $S$. Then by a well-known lemma, $PS$ is a symmedian(isogonal line of the median line, namely, $PM$; the proof is not hard with just trigonometry calculations). Thus we have $\angle APM+\angle BPC=\angle BPS + \angle BPC=180^{\circ}$, as desired(there may be other configurations which does not allow to calculate like this, but here I just wrote based on your figure).

Answer 2

To see this according to classical geometry, put isosceles triangle $ABC$ in a circle, extend $AP$ to $D$, $BP$ to $E$, and join $CE$, $CD$, and $ED$.

Since $\angle PAB=\angle PBC$, then arcs $BD$ and $EC$ are equal, making $CD\parallel EB$.

Again, since $\angle PAC=\angle PBA$, then arcs $CD$ and $AE$ are equal, and $EC\parallel AD$. Thus $PECD$ is a parallelogram.

Now since triangles $PED$ and $PAB$ are similar, and $F$, $M$ are midpoints of $ED$, $AB$$$\frac{AP}{EP}=\frac{AB}{ED}=\frac{AM}{EF}$$Therefore triangles $PFE$ and $PMA$ are similar, with$$\angle APM=\angle EPF$$But$$\angle EPF+\angle BPC=180^o$$Therefore$$\angle APM+\angle BPC=180^o$$