Question

7. In the given figure AB = DE, BF = CE, AB perpendicular to AC and DE
perpendicular to FD. Prove that A BAC = A FDE.
A
C С
B
E
F​

Answer 1

Given :

AB = DE, BF = CE, AB ⟂ AC and DE ⟂ FD

To prove :

∆ BAC ≅ ∆ FDE

Proof :

∵ BF = CE

Adding FC both sides

BF + FC = FC + CE

∴ BC = EF .....(i)

Now, in ∆ BAC and ∆ FDE,

∠A = ∠D = 90°. [∵ AB ⟂ AC and DE ⟂ FD]

Side AB = DE [Given]

Hypotenuse BC = EF [Proved ...(i)]

∴ ∆ BAC ≅ ∆ FDE [By RHS congruence rule]