Sum

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

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#### Solution

In ∆AFD and ∆BFE, we have

∠1 = ∠2 [Vertically opposite angles]

∠3 = ∠4 [Alternate angles]

So, by AA-criterion of similarity, we have

∆FBE ~ ∆FDA

`\Rightarrow \frac{FB}{FD}=\frac{FD}{FA} `

`\Rightarrow \frac{FB}{DF}=\frac{EF}{FA}`

⇒ DF × EF = FB × FA

Concept: Similarity

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