Chapter 5.1, Problem 3PSA

To determine

To Prove: $\angle B\cancel{\cong }\angle CED$ .

Explanation of Solution

Given:

  

  $\begin{array}{l}\overline{AC}\perp \overline{BD}\\ \overline{BC}\cong \overline{EC}\\ \overline{AB}\cancel{\cong }\overline{ED}\end{array}$

Concept Used:

Two angles are said to be congruent if they are of the same measure.

Proof:

Here, we have

  $\begin{array}{l}\text{In }\Delta ABC\text{and }\Delta CED\\ \overline{AC}\perp \overline{BD}\\ \Rightarrow \angle ACB=\angle ECD=90°\\ \overline{BC}\cong \overline{EC}\left(\text{Given}\right)\\ \overline{AB}\cancel{\cong }\overline{ED}\\ \Rightarrow \Delta ABC\cancel{\cong }\Delta CED\\ \Rightarrow \angle B\cancel{\cong }\angle CED\end{array}$