jmc

The area of triangle $ABE$ is $\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}(BE)(20\text{ in.})$. Setting this equal to $100$ square inches we find $BE=10$ inches. The area of triangle $ACD$ is $100-75=25$ square inches. Since triangle $ACD$ is similar to triangle $ABE$ and the ratio of their areas is $\frac{1}{4}$, the ratio of corresponding side lengths is $\sqrt{\frac{1}{4}}=\frac{1}{2}$. Therefore, $CD=\frac{1}{2}BE=\boxed{5}$ inches.

Alternatively, because triangles $ACD$ and $ABE$ are similar, the height-to-base ratio is the same for each of them. In triangle $ABE$, this ratio is $\frac{20\text{ in.}}{10\text{ in.}}=2$. Therefore, the height of $ACD$ is $2\cdot CD$. Solving $\frac{1}{2}(2\cdot CD)(CD)=25{\text{ in.}}^2$ we find $CD=5$ inches.