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

Answer 1

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