Proving angles are supplementary in isosceles triangle Let $ABC$ be a triangle with $AC=BC$, and let $P$ be a point inside $\triangle ABC$, satisfying $\angle PAB=\angle PBC$. If $M$ is the midpoint of $AB$, show that $\angle APM+\angle BPC=180^{\circ}$. 
Here is what I tried:
Let $\angle PAB=\angle PBC=x$. Let $\angle CAB=\angle CBA=A\implies\angle CAP=\angle ABP=A-x$ Working backwards, if $\angle APM+\angle BPC=180^{\circ}$, then $\sin\angle APM=\sin\angle BPC$. By Sine Law on $\triangle BPC,\triangle APM$, $\sin\angle APM=\frac{AM\sin x}{PM},\sin\angle BPC=\frac{BC\sin x}{PC}$. Setting these equal and canceling, we get $\frac{AM}{PM}=\frac{BC}{PC}$, so if we can prove that, then we are done. I tried plugging in $BM$ for $AM$ (since $M$ is the midpoint), which gives $\frac{BM}{PM}=\frac{BC}{PC}$ and can be rearranged to $\frac{BM}{BC}=\frac{PM}{PC}$. Let $BP\cap CM=Q$. Using Sine Law again on $\triangle BCM,\triangle BMQ, \triangle BCQ$, we can write $\frac{BM}{BC}=\frac{PM}{PC}\implies\frac{QM\sin x}{QC\sin(A-x)}=\frac{QM\sin\angle QAC}{QC\sin\angle QPM}$, so $\frac{\sin x}{\sin(A-x)}=\frac{\sin\angle QAC}{\sin\angle QPM}$. If any of the ratios I got could be proven then we would be done.
Also, I found this problem in a handout on the symmedian lemma: In $\triangle ABC$, let the intersection of the tangents to the circumcircle through $B,C$ respectively be $D$. Then $AD$ coincides with the $A$-symmedian. I was wondering if that leads to a shorter solution, but I think my law of sines solution eventually leads to something provable.
 A: SOLUTION WITHOUT USING LAW OF SINES
Note that if $CP$ met $AB$ on point $T$, you need to prove that $\angle BPM = \angle APT$. This implies that $PT$ is the $P$-symmedian of $APB$. 
Thus, we must prove that $BT:AT=BP^2:AP^2$.
Draw the excircle of $\triangle APB$, $\Gamma$. 
Note that $BC$, $AC$ are lines tangent to $\Gamma$. 
If $CT$ meets $\Gamma$ on point $K(\neq P)$, Note that $\sin PBK= \sin PAK$. 
Using this, notice we must prove that $BK:AK=BP:AP$. 
Since $\triangle BPC$ and $\triangle KBC$ are similar, notice that $BK:CK=BP:BC$. 
In a similar fashion, $AK:CK=PA:CA=PA:BC$. 
This implies  $BK:AK=BP:AP$, thus we get our desired result. 
A: Indeed, we just need to prove that $PD$ is the $P$-symmedian of $APB$, where $D=CP \cap AB$.
We prove $\frac{PA^2}{PB^2}=\frac{DA}{DB}$.
Let $\angle PAB=\angle PBC=x$, $\angle PAC=\angle PBA=y$.
From Sine Law on $PAB$, we get $\frac{PA}{PB}=\frac{\sin y}{\sin x}$.
From Trig Ceva, we have $$\frac{\sin^2 x \sin \angle PCA}{\sin^2 y \sin \angle PCB}=1$$
Now, we have $$\frac{DA}{DB}=\frac{\triangle CAD}{\triangle CDB} = \frac{CA \sin \angle ACD}{CB \sin \angle BCD}=\frac{\sin^2 y}{\sin^2 x}=\left(\frac{PA}{PB}\right)^2$$ which is enough to claim that $PD$ is the $P$-symmedian of $APB$.
