3. As in Figure 1, in quadrilateral $A B C D$, it is known that
$$
\begin{array}{l}
\text { } \angle A C B=\angle B A D \\
=105^{\circ}, \angle A B C= \\
\angle A D C=45^{\circ} \text {. Then } \\
\angle C A D=(\quad) .
\end{array}
$$
(A) $65^{\circ}$
(B) $70^{\circ}$
(C) $75^{\circ}$
(D) $80^{\circ}$
3. C.

It is easy to know that, $\angle C A D=105^{\circ}-\left(180^{\circ}-105^{\circ}-45^{\circ}\right)=75^{\circ}$.