Chapter 3.7, Problem 4PSA

To determine

To Prove: $\Delta EBC$ is isosceles

Explanation of Solution

Given information: $\angle 3\cong \angle 6$ , $\angle 3$ is complement to $\angle 4$ and $\angle 6$ is complement to $\angle 5$ is provided in the question which will help to find the solution.

Proof: Here, $\angle 3+\angle 4={90}^{\circ }$ and $\angle 6+\angle 5={90}^{\circ }$ (given)

  →$\angle 3+\angle 4=\angle 6+\angle 5$

  →$\angle 3+\angle 4=\angle 3+\angle 5$

  →$\angle 4=\angle 5$

So that, $\angle 4\cong \angle 5$ (by CPCT)

So, BE = EC (equal angles of equal sides)

Hence, we can say that

  $\Delta EBC$ is isosceles.