1. Given a convex pentagon $A B C D E, \angle A B C=\angle A E D=$ $90^{\circ}, \angle D A C=30^{\circ}, \angle B A E=70^{\circ}, F$ is the midpoint of side $C D$, and $F B=F E$. Then $\angle B A C$ equals ( ).
(A) $10^{\circ}$
(B) $20^{\circ}$
(C) $30^{\circ}$
(D) $15^{\circ}$
$-1 . \mathrm{B}$
Take the midpoints $P, Q$ of $A C, A D$, and connect $P B, P F$, $Q E, Q F$, then $P B=P A$ $=F Q, Q E=Q A=$ $F P$. Furthermore, since $F B=F E$, we have $\triangle F P B \cong \triangle F Q E$. Thus, $\angle 17 \angle 3=\angle 2+\angle 4$. Also, $\angle 1=\angle 2=\angle C A D$, so $\angle 3=\angle 4$, and $\angle 3=2 \angle B A C$, $\angle 4=2 \angle Q A E$, hence, $\angle B A C=\angle Q A E$. Given that $\angle B A E$ $=70^{\circ}, \angle D A C=30^{\circ}$, then $\angle B A C=20^{\circ}$.