1996 AMC 12 第 21 题

Let $\angle BAC=\alpha$ and express $\angle C$ using isosceles triangle $ABC$

Use $BD\perp AC$ to find the vertex angle at $B$ of isosceles triangle $ABD$

Let $\angle BAC=\alpha.$ Since $AB=AC,$ $$\angle C=\frac{180^\circ-\alpha}{2}=90^\circ-\frac{\alpha}{2}.$$ Because $BD\perp AC,$ the angle between $BA$ and $BD$ is $90^\circ-\alpha.$ In isosceles triangle $ABD,$ $AB=BD,$ so its two base angles are $$\begin{aligned} \angle D &=\frac{180^\circ-(90^\circ-\alpha)}2\\ &=45^\circ+\frac{\alpha}{2}. \end{aligned}$$ Their sum is $135^\circ,$ so the correct answer is D.