Chapter 7.4, Problem 9PSB

To determine

To prove : $\Delta AEF$ is isosceles

Explanation of Solution

Given information : The following information has been given

  $\overrightarrow{AB}\cong \overrightarrow{AD}$

  $FC\perp BD$

Formula used : The summation of the angles of a triangle will be ${180}^0$

Proof : We know that $\overrightarrow{AB}\cong \overrightarrow{AD}$ , hence we can say that

  $\angle B\cong \angle D=x$

Now, we assume that

  $\angle DEC=a$

Thus, $\angle FEA=a$ (vertically opposite)

Hence, in triangle AFE, we add up the angles and equate it to the exterior angle EAB.

  $x+(90-a)=x+a$

Hence, we get

  $a={45}^0=x$

Thus, two angles of triangle AEF are equal, hence it is an isosceles triangle.