Q24 of 56 Page 6

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

                                                                                                                                


Given: The diagonal BD of parallelogram ABCD intersects the segment AE at F, where E is any point on BC.
To prove: DF × EF = FB × FA
Proof: In triangles AFD and BFE,
∠FAD = ∠FEB (Alternate angles)
∠AFD = ∠BFE (Vertically opposite angles)
Therefore ΔADF ∼ ΔBFE (AA similarity)

Hence DF × EF = FB × FA



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22 In Δ ABC, ∠ B = 90° and D is the mid-point of BC. Prove that AC2 = AD2 + 3CD2.
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