In the given figure, AB ⊥ BC, GF ⊥ BC and DE ⊥ AC. Prove that ΔADE ~ Δ GCF.
Given:
AB
BC
GF
BC
DE
AC
Since AB
BC so ∠DAE & ∠GCF are complementary angles i.e.
∠DAE + ∠GCF = 900 ….Equation 1
Similarly since GF
BC 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
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