The diagonal BD of parallelogram ABCD intersects AE at 'F'. 'E' is any point on BC.
Prove that DE.EF = FB. FA

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Solution

$A B C D$ is a parallelogram.

In $△ A D F$ and $△ E F B$ ,,

$∠ A D F = ∠ E B F$ (Alternate angles)
$∠ D A F = ∠ B E F$ (Alternate angles)

$∠ A F D = ∠ B F E$ (Vertically opposite angles)
Therefore, $△ A D F \sim△ E F B$ and thus,

$\frac{A F}{E F} = \frac{D F}{F B} \Rightarrow A F \times F B = D F \times E F$
Hence $A F \times F B = D F \times E F$ .

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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.

B D

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D F \times E F = F B \times F A

In quadrilateral ABCD, diagonals AC and BD intersect at point E. Such that AE : EC = BE : ED. Then, ABCD is a parallelogram.

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