Question

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Answer 1

Answer:

Step-by-step explanation:

In Δ AGF and Δ DBG,

∠ AGF = ∠ GBD (Corresponding angles)  

∠ GAF = ∠ BDG = 90° each

∴ Δ AGF ~ ΔDBG ....

Similarly, Δ AFG ~ Δ ECF (AA similarity) ... ),

we get 

Δ DBG ~ Δ ECF

⇒ BD/EF = BG/FC = DG/EC

BD/EF = DG/EC

EF × DG = BD × EC ....1

Also DEFG is a square ⇒ DE = EF = FG = DG ......2

From (1) and (2) we get 

DE² = BD × EC

Hence proved.