Question

□DEFG is a square and ∠BAC = 90º. Provethat : (a) ∆ AGF ~ ∆DBG (b) ∆AGF ~ ∆EFC (c) ∆DBG ~ ∆EFC (d) DE2 = BD × EC

Answer 1

Solution:-
Given ; ABC is a triangle in which ∠ BAC = 90° and DEFG is a square.
Proof in : (1) In Δ AGF and Δ DBG,
∠ AGF = ∠ GBD    (Corresponding angles)
∠ GAF = ∠ BDG = 90° each
So, Δ AGF ~ Δ DBG (Proved By AA similarity)
(2) In Δ AGF and Δ EFC,
∠ AFG = ∠ FCE   (Corresponding angles)
∠ GAF = ∠ CEF = 90° each
So, Δ AGF ~ Δ EFC  (Proved by AA similarity)
(3) In Δ DBG and Δ EFC,
∠ DBG = ∠ ECF = (Corresponding angles)
∠ BDG = ∠ CEF = 90° each
So, Δ DBG ~ Δ EFC  (Proved by AA similarity)
(4) In Δ AGF and Δ DBG,
∠ AGF = ∠ GBD (Corresponding angles)  
∠ GAF = ∠ BDG = 90° each
∴ Δ AGF ~ ΔDBG .....(1)
Similarly, Δ AFG ~ Δ ECF (AA similarity) ....(2)
From (1) and (2), we get 
Δ DBG ~ Δ ECF
⇒ BD/EF = BG/FC = DG/EC
BD/EF = DG/EC
EF × DG = BD × EC ....(3)
Also DEFG is a square ⇒ DE = EF = FG = DG ....(4)
From (3) and (4), we get 
DE² = BD × EC
Hence proved.

Answer 2

I hope this will help you..!
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