Question

ABCD is a parallelogram and E is a point on BC. If the diagonal BD intersect AE at F, prove that AF×FB =EF × FD

Answer 1

its soo simple buddy !!!

Given: The diagonal BO of parallelogarn ABCD intersects the segment AE at F,

where E is any point on BC.

To provo: DF x EF= FB x FA

Proof: In triangles AFD and BFE,

∠FAD = ∠FEB (Alternate angles)

∠AFD = ∠BFE (Vertically opposite angles)

Therefore △ADF ~ △BFE (AA similarity)

DF/FA = FB/EF

Hence DF x EF = FB x FA