Question

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

Solution

1

Parallelogram ABCD with diagonal BD intersecting line segment AE at point F

+1 mark

2

In $△ADF$ and $△EBF$: $\angle DFA=\angle EFB$ (vertically opposite angles) and $\angle ADF=\angle FBE$ (alternate angles, since $AD\parallel BC$)

+2 marks

3

$\therefore △ADF\sim △EBF$ (AA similarity). So $\frac{DF}{FB}=\frac{FA}{EF}$

+1 mark

4

$\Rightarrow DF\times EF=FB\times FA$

+1 mark

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