Question

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

Answer 1

but where the points show in figure ABC

Answer 2

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.