5. (10 points) As shown in the figure, $B H$ is the height of the right trapezoid $A B C D$, and $E$ is a point on the diagonal $A C$ of the trapezoid. If the areas of $\triangle D E H$, $\triangle B E H$, and $\triangle B C H$ are $56$, $50$, and $40$ respectively, then the area of $\triangle C E H$ is ( )
A. 32
B. 34
C. 35
D. 36
【Analysis】As shown in the figure below: Draw $E F \perp D C, E G \perp B H$ through point $E$, and connect $A F, B F, B D$. By the principle that triangles with equal bases and heights have equal areas, we get $S_{\triangle B D F}=S_{\triangle A D F}, S_{\triangle A D C}=S_{\triangle B D C}$. Therefore, we have: $S_{\triangle C D E}=S_{\triangle A D C}-S$ $\triangle A D E=S_{\triangle B D C}-S_{\triangle B D F}=S_{\triangle B F C}$. And $S_{\triangle B F C}=S_{\triangle B F H}+S_{\triangle B C H}=S_{\triangle B E H}+S_{\triangle B C H}=90$; thus $S_{\triangle C H E}$ $=S_{\triangle E D C}-S_{\triangle H D E}=90-56=34$. Based on this, the problem can be solved.
【Solution】Solution: As shown in the figure above, draw $E F \perp D C, E G \perp B H$ through point $E$, and connect $A F, B F, B D$, then $S_{\triangle B D F}=S_{\triangle A D F}, S_{\triangle A D C}=S_{\triangle B D C}$,

so $S_{\triangle C D E}=S_{\triangle A D C}-S_{\triangle A D E}=S_{\triangle B D C}-S_{\triangle B D F}=S_{\triangle B F C}$,
and since $S_{\triangle B F C}=S_{\triangle B F H}+S_{\triangle B C H}=S_{\triangle B E H}+S_{\triangle B C H}=90$,
thus $S_{\triangle C H E}=S_{\triangle E D C}-S_{\triangle H D E}=90-56=34$.
Therefore, the answer is: $B$.