Question

In the following figure, DEFG is a square in a triangle ABC right angled at A.

Prove that

1. ΔAGF ∼ ΔDBG
2. ΔAGF ∼ ΔEFC

Answer 1

Given, ABC is a triangle in which ∠BAC = 90° and DEFG is a square.

i. In ΔAGF and ΔDBG,

∠AGF = ∠GBD   ...(Corresponding angles)

∠GAF = ∠BDG   ...(Each 90°)

So, ΔAGF ∼ ΔDBG   ...(By AA similarity)

Hence Proved.

ii. In ΔAGF and ΔEFC,

∠AFG = ∠FCE   ...(Corresponding angles)

∠GAF = ∠CEF   ...(Each 90°)

So, ΔAGF ∼ ΔEFC   ...(By AA similarity)

Hence Proved.