Question

Please solve this question..​

Answer 1

Step-by-step explanation:

(i)

In △AGF and △DBG,

∠AGF = ∠GBD (corresponding angles)

∠GAF = ∠BDG (each = 90‘)

∴△AGF ~ △DBG

(ii)

Similarly, △AFG ~ △ECF (AA Similarity)

(iii)

From (i) and (ii), △DBG ~ △ECF

BD/EF - BG/FC - DG/EC

BD/EF - DG/EC

EF × DG = BD × EC

Also DEFG is a square ⇒ DE = EF = FG = DG - (iv)

From (iii) and (iv),

DE^2 = BD × EC.

Hope it helps!