1994 AMC 12 第 13 题

Let $\angle A=\theta$ and use $AP=PC$ in $\triangle APC$

Use $PC=CB$ to relate $\angle PCB$ to a base angle of $\triangle ABC$

Let $\angle A=\theta.$ Since $AP=PC,$ triangle $APC$ has $\angle PAC=\angle PCA=\theta.$ Since $AB=AC,$ each base angle of $\triangle ABC$ is $\frac{180^\circ-\theta}{2}.$ Also $PC=CB,$ so the base angles of $\triangle PCB$ at $P$ and $B$ are equal; the one at $B$ is $\frac{180^\circ-\theta}{2}.$ Hence $\angle PCB=\theta.$ At $C,$ $$2\theta=\frac{180^\circ-\theta}{2},$$ giving $5\theta=180^\circ$ and $\theta=36^\circ.$ Thus the correct answer is B.