Question

Given: FB = FD, AE ⊥ FD and FC ⊥ AD.

Prove that: `(FB)/(AD) = (BC)/(ED)`.

Answer 1

Given, FB = FD

∴ ∠FDB = ∠FBD   ...(1)

In ΔAED = ΔFCB,

∠AED = ∠FCB = 90°

∠ADE = ∠FBC  ...[Using (1)]

ΔAED ~ ΔFCB  ...[By AA similarity]

∴ `(AD)/(FB) = (ED)/(BC)`

`(FB)/(AD) = (BC)/(ED)`