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Question
The diagonal BD of a parallelogram ABCD intersects the segment AE at the point F, where E is any point on the side BC. Prove that DF × EF = FB × FA
Solution
In ∆AFD and ∆BFE, we have
∠1 = ∠2 [Vertically opposite angles]
∠3 = ∠4 [Alternate angles]
AD || BC
So, by AA-criterion of similarity, we have
∆FBE ~ ∆FDA
⇒ `(FB)/(FD) = (FD)/(FA)`
⇒ `(FB)/(DF)= (EF)/(FA)`
⇒ DF × EF = FB × FA
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