Question

the diagonal BD of a parallelogram ABCD intersect the line segment AE at the point F, where E is any point on the side BC prove that DF× EF=FB×FA.

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

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