$17 \cdot 33$ In $\triangle A B C$, $\angle A=70^{\circ}$, $D$ is a point on side $A C$, the angle bisector $A E$ of $\angle A$ intersects $B D$ at $H$, and $A H: H E=3: 1$, also $B H: H D=5: 3$. Then the degree measure of $\angle C$ is
(A) $45^{\circ}$.
(B) $55^{\circ}$.
(C) $75^{\circ}$.
(D) $80^{\circ}$.
(Hanjiang Cup Junior Mathematics Competition, 1991)
[Solution] Connect $CH$, as shown in the figure, let the areas of each triangle be $S_{0}, S_{1}, S_{2}, S_{3}, S_{4}$, and set $S_{0}=1$.
From $A H: H E=3: 1$, we know $S_{1}=3$.
Also, $B H: H D=5: 3$, then
$$
S_{2}=\frac{3}{5} S_{1}=\frac{9}{5} \text {. }
$$

At the same time, $\left(S_{2}+S_{3}\right): S_{4}=3: 1$,
then $S_{4}=\frac{1}{3}\left(S_{2}+S_{3}\right)$.
Furthermore, $S_{3}:\left(S_{4}+S_{0}\right)=S_{3}:\left[\frac{1}{3}\left(S_{2}+S_{3}\right)+S_{0}\right]=3: 5$, i.e., $S_{3}:\left[\frac{1}{3}\left(\frac{9}{5}+S_{3}\right)+1\right]=3: 5$, solving this gives $S_{3}=\frac{6}{5}$,
then
$$
S_{4}=\frac{1}{3}\left(S_{2}+S_{3}\right)=\frac{1}{3}\left(\frac{9}{5}+\frac{6}{5}\right)=1 .
$$

Thus, $S_{0}+S_{1}=4$, and $S_{2}+S_{3}+S_{4}=\frac{9}{5}+\frac{6}{5}+1=4$,
i.e., $S_{\triangle A C E}=S_{\triangle A B E}$, which means $C E=E B$.
Also, $A E$ bisects $\angle A$, so $\triangle A B C$ is an isosceles triangle.
$$
\therefore \angle C=\frac{1}{2}\left(180^{\circ}-\angle \mathrm{A}\right)=\frac{1}{2}\left(180^{\circ}-70^{\circ}\right)=55^{\circ} \text {. }
$$

Therefore, the answer is $(B)$.