Chapter 1.6, Problem 10PSC

To determine

To Prove: $\angle CAE\text{ is a right angle}\text{.}$

Explanation of Solution

Given:

  

  $\begin{array}{l}\overrightarrow{AC}\text{ bisects }\angle BAD\\ \overrightarrow{AE}\text{ bisects }\angle DAF\end{array}$

Concept Used:

Sum of linear pairs of angles is $180°$ . Bisectors divides the angle into two equal parts.

Proof:

Here, we have

  $\begin{array}{l}\overrightarrow{AC}\text{ bisects }\angle BAD\\ \Rightarrow \angle BAC=\angle CAD\\ \overrightarrow{AE}\text{ bisects }\angle DAF\\ \Rightarrow \angle DAE=\angle EAF\end{array}$

Also,

  $\begin{array}{l}\angle BAD+\angle DAF=180°\\ \Rightarrow \frac{1}{2}\times \left(\angle BAD+\angle DAF\right)=\frac{1}{2}\times 180°\\ \Rightarrow \frac{1}{2}\angle BAD+\frac{1}{2}\angle DAF=90°\\ \Rightarrow \angle CAD+\angle DAE=90°\\ \Rightarrow \angle CAE=90°\\ \Rightarrow \angle CAE\text{ is a right angle}\text{.}\end{array}$