23 、As shown in the figure, draw $D O$ to bisect $\angle A D C$, and draw $B O$ to bisect $\angle A B C$, $B O$ intersects $D O$ at point $\mathrm{O}$. Given $\angle \mathrm{A}=35^{\circ}, \angle \mathrm{O}=42^{\circ}$. Then $\angle \mathrm{C}=$ _ . $\qquad$
【Analysis】As shown in the figure, connect BD.
By the sum of the interior angles of a triangle being $180^{\circ}$, we have
$$
\begin{array}{l}
\angle C D B+\angle D B C+\angle C=180^{\circ} \Rightarrow 2 \angle O D A+\angle A D B+\angle D B C+\angle C=180^{\circ} \\
\angle A B D+\angle A D B+\angle A=180^{\circ} \Rightarrow 2 \angle O B C+\angle C B D+\angle A D B+\angle A=180^{\circ} \\
\angle O D B+\angle O B D+\angle O=180^{\circ} \\
\Rightarrow 2(\angle O D A+\angle A D B)+2(\angle O B C+\angle C B D)+2 \angle O=360^{\circ}
\end{array}
$$

Therefore,
$$
\begin{array}{l}
2 \angle O D A+\angle A D B+\angle D B C+\angle C+2 \angle O B C+\angle C B D+\angle A D B+\angle A \\
=2(\angle O D A+\angle A D B)+2(\angle O B C+\angle C B D)+2 \angle O
\end{array}
$$

So $\angle A+\angle C=2 \angle O \Rightarrow \angle C=49^{\circ}$