In triangle $ABC$, $AB = AC$. Point $P$ lies on $\overline{AB}$ such that $CP = BC$. If $\angle APC = 115^\circ$, what is $\angle ACP$ (in degrees)?
A
P 115
B C
Angle BPC is supplemental to APC = 65°
In triangle BPC, since CP = BC then angle PBC = 65°
Then angle PCB = 180 - 2 (65) = 50°
But AB = AC....so angle ABC = 65° = angle ACB
And
ACB = PCB + ACP
65 = 50 + ACP
65 - 50 = ACP = 15°
+195 
