Chapter 2.2, Problem 7PSA

To determine

To Prove: $\angle CDE$ is supplementary to $\angle FDE$

Explanation of Solution

Given information: $CD\perp BE$

  

Proof:

  $\begin{array}{l}CD\perp BE\mathrm{....}(given)\\ \therefore \angle CDE={90}^{\circ }\mathrm{....}(CD\perp BE).\\ \therefore \angle CDF+\angle FDE={90}^{\circ }\mathrm{....}(\because \angle CDE=\angle CDF+\angle FDE).\end{array}$

Hence $\angle CDF$ and $\angle FDE$ are complementary.