10. (10 points) As shown in the figure, $A B / / C E, A C / / D E$, and $A B=A C=5, C E=D E=10$. If the area of triangle $C O D$ is 10, find the area of quadrilateral $A B D E$.


【Analysis】Based on “$A B / / C E, A C / / D E$”, deduce the relationship between the ratio of the areas of triangles and the ratio of the sides, and find out each.

【Solution】Solution: Since $A C / / D E$, we have $S \triangle_{A O E}=S_{\triangle C O D}$,
$$
\text{Also, } \frac{O C}{C E}=\frac{S_{\triangle C O D}}{S_{\triangle C D E}}, \frac{O E}{C E}=\frac{S \triangle_{A O E}}{S_{\triangle E A C}}=\frac{S \triangle_{C O D}}{S_{\triangle E A C}},
$$

Therefore, $\frac{O C}{O E}=\frac{S \triangle E A C}{S \triangle_{C D E}}$.
Since the height of triangle $E A C$ on side $A C$ is equal to the height of triangle $C D E$ on side $D E$,
we have $\frac{O C}{O E}=\frac{S E A C}{S \triangle_{C D E}}=\frac{A C}{D E}=\frac{1}{2}$.
Since $\frac{S_{\triangle C O D}}{S_{\triangle C D E}}=\frac{O C}{O E}=\frac{1}{2}$, we have $S \triangle D O E=2 S \triangle C O D=20$.
Since $\frac{S \triangle_{A O C}}{S \triangle_{\mathrm{AOE}}}=\frac{O C}{O E}=\frac{1}{2}$, we have $S_{\triangle A O C}=\frac{1}{2} S_{\triangle A O E}=\frac{1}{2} S \triangle C O D=5$.
Therefore, $S_{\triangle A C E}=S \triangle A O C+S \triangle_{A O E}=15$,
Since $A B / / C E$, we have $\frac{S \triangle_{\mathrm{ABC}}}{\mathrm{S} \triangle_{\mathrm{ACE}}}=\frac{\mathrm{AB}}{\mathrm{CE}}=\frac{1}{2}$,
which means $S_{\triangle A B C}=\frac{1}{2} S_{\triangle A C E}=\frac{1}{2} \times 15=7.5$.
Therefore, $S_{A B C D E}=S_{\triangle A B C+S \triangle} A C E+S_{\triangle C O D}+S \triangle D O E=7.5+15+10+20=52.5$.
Answer: The area of quadrilateral $A B D E$ is 52.5.