Chapter 4.1, Problem 18PSC

To determine

To Prove: $\Delta AFB$ is isosceles

Explanation of Solution

Given information: $AD=DB,AE=BC,CD=ED.$

Formula Used: Congruency of triangles.

Proof:

  $\begin{array}{l}AD=DB,AE=BC,CD=ED.\left(\text{Given}\right)\\ \therefore \Delta AED\cong \Delta BCD\text{(SSS)}\\ \therefore \angle EAD=\angle CBD\text{(CPCT)}\\ \angle DAB=\angle DBA\left(\text{Angles opposite to equal sides are equal}\right)\\ \therefore \angle EAD+\angle DAB=\angle CBD+\angle DBA\\ \therefore \angle EAB=\angle CBA\\ \therefore \Delta EAB\cong \Delta CBA(\text{SAS)}\\ \therefore BE=AC(\text{CPCT})\\ \therefore \Delta BED\cong \Delta DAC\text{(SSS})\\ \therefore \angle DEC=\angle EBD(\text{CPCT})\\ \therefore \angle FAB=\angle FBA\left(\text{Subtraction property}\right)\\ \Delta FAB\text{is isosceles}\text{.}\end{array}$