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Question
Given: FB = FD, AE ⊥ FD and FC ⊥ AD.
Prove that: `(FB)/(AD) = (BC)/(ED)`.
Solution
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)`
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