In the given figure, AB ⊥ BC, GF ⊥ BC and DE ⊥ AC. Prove that ΔADE ~ Δ GCF.

Given:

AB BC

GFBC

DE AC

Since AB BC so ∠DAE & ∠GCF are complementary angles i.e.

∠DAE + ∠GCF = 90⁰ ….Equation 1

Similarly since GFBC so ∠CFG & ∠GCF are complementary angles i.e.

∠CGF + ∠GCF = 90⁰ ….Equation 2

Combining Equation 1 & 2 We can say that

∠CGF = ∠DAE

Also ∠CFG = ∠DEA (Perpendicular Angles)

So ΔCGF is similar to ΔADE By A.A. (Angle Angle)axiom of similarity

Hence Proved