Question

the diagonal BD of a parallelogram ABCD intersect the segment a at the point if there is any point on the side BC prove that DF into f is equal to FB into FA

Answer 1

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

Hope This helps :)

Answer 2

Step-by-step explanation:

In ΔAFD and ΔBFE, we have

∠1 = ∠2 (Vertically opposite angles)

∠3 = ∠4 (Alternate interior angles)

∴ ΔFBE ≈ ΔFDA (AA similarity)

So, FB / FD = FE / FA

By cross multiplication, we get

 DF * EF = FB * AF

Hence proved.

Hope it helps.