Question

In the following figure, AB =EF, BC = DE and $\angle$ B = $\angle$ E = ${90}^0$.

Prove that AD =FC.

Answer 1

In $\Delta$ ABD and $\Delta$ FEC

AB=FE ( given )

$\angle FEC=\angle ABD({90}^0$

BC = DE
in which CD is common part coming in both triangles
BC + CD = CD + DE
$\to BD=CE$
$\therefore \Delta ABDis\cong \Delta FEC$ by SAS rule of congruence

so AD=FC