3. In quadrilateral $ABCD$, $AD \parallel BC, \angle ABC$ $=80^{\circ}, AB=AD=\frac{1}{2} BC, CH \perp AB$ at point $H$, and connect $DH$. Then $\angle CHD=(\quad)$.
(A) $30^{\circ}$
(B) $35^{\circ}$
(C) $40^{\circ}$
(D) $45^{\circ}$
3. C.

As shown in Figure 5, take the midpoint $E$ of $B C$, and connect $H E$ and $D E$. It is easy to see that $A D \Perp B E$.
Then $H E=B E$
$$
=A B=D E \text {, }
$$

and $D E \perp C H$.
Therefore, $\angle C H D=90^{\circ}-\angle H D E$
$$
=\frac{1}{2} \angle H E D=\frac{1}{2} \angle H B E=40^{\circ} \text {. }
$$