17・28 $\triangle A B C$ and $\triangle A B D$ overlap (as shown in the figure), and $A B=A C=B D$, also $B D$ and $A C$ intersect at $E$ and $B D \perp A C$, then $\angle C+\angle D=$
(A) $115^{\circ}$
(B) $120^{\circ}$
(C) $130^{\circ}$
(D) $135^{\circ}$
(E) uncertain.
[Solution] In $\triangle A B C$ and $\triangle A B D$, given $A B=A C=B D$,
$$
\begin{array}{l}
\text { then } \angle C=\frac{180^{\circ}-\angle B A C}{2}, \angle D=\frac{180^{\circ}-\angle D B A}{2}, \\
\therefore \angle C+\angle D=180^{\circ}-\frac{1}{2}(\angle B A C+\angle D B A), \\
\text { and } \angle B A C+\angle D B A=90^{\circ}, \\
\therefore \angle C+\angle D=180^{\circ}-45^{\circ}=135^{\circ} .
\end{array}
$$

Therefore, the answer is $(D)$.