13. As shown in Figure 2, in

isosceles $\triangle ABC$, $AC = BC$, $\angle ACB = 40^{\circ}$. A circle with $BC$ as its diameter intersects $AC$ and $AB$ at points $D$ and $E$ respectively, and the diagonals of quadrilateral $BCDE$ intersect at point $F$. Then $\angle BFC=$ ( ).
(A) $90^{\circ}$
(B) $100^{\circ}$
(C) $105^{\circ}$
(D) $110^{\circ}$
(E) $120^{\circ}$
13. D.

From the problem, we know
$$
\begin{array}{l}
\angle B E C=\angle B D C=90^{\circ}, \\
\angle C B D=50^{\circ}, \\
\angle B C E=\frac{1}{2} \angle A C B=20^{\circ} .
\end{array}
$$

Therefore, $\angle B F C=180^{\circ}-50^{\circ}-20^{\circ}=110^{\circ}$