# In the adjoining figure, the diagonal BD of a parallelogram ABCD intersects the segment AE at the point F, where E is any point on the side BC. Show that DF x FE = BF x FA.

Given: ABCD is a parallelogram

To Prove: DF x FE = BF x FA

In AFD and BFE

1 = 2 (alternate angles)

3 = 4 (vertically opposite angles)

AFD ~BFE (by AA similarity criterion)

So,

(corresponding sides of similar triangle are proportional)

DF x FE = BF x FA

Hence Proved

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