2. As shown in Figure 1, in
$\triangle A B C$, $\angle A B D$
$=\angle D B E=\angle E B C$,
$\angle A C D=\angle D C E=$
$\angle E C B$. If $\angle B E C$
$=145^{\circ}$, then $\angle B D C$
is equal to ( ).
$\begin{array}{ll}\text { (A) } 100^{\circ} & \text { (B) } 105^{\circ}\end{array}$
(C) $110^{\circ}$
(D) $115^{\circ}$
2.C.

From $\frac{1}{3}(\angle B+\angle C)=180^{\circ}-145^{\circ}=35^{\circ}$, we have $\frac{2}{3}(\angle B+\angle C)=70^{\circ}$.
Then $\angle B D C=180^{\circ}-70^{\circ}=110^{\circ}$.