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Answer:
hence proved, I hope you understood easil.
Answer:
In Δ AGF and Δ DBG
∠ A = ∠ BDG ( each 90° )
∠ AGF = ∠ DBG ( Corresponding angles )
∴ Δ AGF ~ Δ DBG ( A - A test of similarity ) _ _ _ ( I )
In Δ AGF and Δ EFC
∠ AFG = ∠ ECF ( corresponding angles )
∠ A = ∠ CEF ( each 90° )
∴ Δ AGF ~ Δ EFC ( A - A test of similarity ) _ _ _ ( II )
From ( I ) & ( II )
Δ DBG ~ Δ EFC
∴ DB / EF = DG / EC ( c.s.s.t. )
∴ BD / DE = DE / EC ( ∴ DE = EF )
∴DE² = BD × EC
hope it helps you
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