Question

In the following figure, AB = EF, BC = DE and ∠B = ∠E = 90°.

Prove that AD = FC.

Answer 1

Given that, BC = DE
⇒ BC + CD = DE + CD  ....( Adding CD on both sides )
⇒ BD = CE                     ....(i)
Now, in ΔABD and ΔFEC,
AB = EF                         ....(given)
∠ABD = ∠FEC               ....(Each 90°)
BD = CE                        ....[ From (i) ]
⇒  ΔABD ≅  ΔFEC         ...(by SAS congruence criterion)
⇒ AD = FC                    ...(c.p.c.t.)