Question

the diagonals BD of parallelogram ABCD intersect AE at F .E
any point on BC prove that DF.EF. is equal to FB.FA

Answer 1

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

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