Chapter 2.5, Problem 10PSA

To determine

To Prove: $\angle BAC\cong \angle EAD$

Answer to Problem 10PSA

  $\angle BAC\cong \angle EAD$

Explanation of Solution

Given information: $\angle BAD$is right angle, $CA\perp AE$ .

  

Proof: $\angle BAD$ is right angle (given)

  $\begin{array}{l}\therefore \angle BAD={90}^{\circ }\\ CA\perp AE\mathrm{....}(given)\\ \therefore \angle CAE={90}^{\circ }\end{array}$

  $\therefore \angle BAD=\angle CAE$ (Each equal ${90}^{\circ }$ )

  $\therefore \angle BAC+\angle CAD=\angle CAD+\angle DAE$

  $\begin{array}{l}\therefore \angle BAC+\angle CAD=\angle CAD+\angle DAE\\ \therefore \angle BAC=\angle DAE\end{array}$