In triangle ABC, AB = AC. Point P lies on line AB such that CP = BC. If angle APC = 115 degrees, what is angle ACP(in degrees)?
So far I have figured out that the answer is not 65 or 70.
\(\angle APC = 115^\circ\\ \angle BPC = 65^\circ\\ \because CP = CB \\ \therefore \angle CBP = 65^\circ\\ \because AB = AC\\ \angle ACB = 65^\circ\\ \)
Then you consider △CPB to find angle PCB. The rest is trivial.
