Chapter 1.6, Problem 9PSB

To determine

To Prove: $\angle A\cong \angle BCD$

Explanation of Solution

Given:

  

  $\begin{array}{l}\overrightarrow{CE}\text{ bisects }\angle BCD\\ \angle A\text{is a right angle}\text{.}\\ m\angle BCE=45°\end{array}$

Concept Used:

Bisectors divides the angle into two equal parts.

Concept Used:

Sum of linear pairs of angles is $180°$ .

Proof:

Here, we have

  $\begin{array}{l}\overrightarrow{CE}\text{ bisects }\angle BCD\\ m\angle BCE=m\angle DCE=\frac{1}{2}\left(\angle BCD\right)\\ \because m\angle BCE=45°\\ \Rightarrow m\angle BCD=2\left(m\angle BCE\right)\\ \Rightarrow m\angle BCD=2\times 45°\\ \Rightarrow m\angle BCD=90°\left(1\right)\\ \because \angle A\text{is a right angle}\\ \Rightarrow m\angle A=90°\left(2\right)\\ \text{From }\left(1\right)\text{and}\left(2\right)\text{, we get }\\ \angle A\cong \angle BCD\end{array}$