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. - Zigya
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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.


In ∆AFD and ∆BFE, we have
∠1 = ∠2 [Vertically opposite angles]
∠3 = ∠4 [Alternate angles]

In ∆AFD and ∆BFE, we have∠1 = ∠2 [Vertically opposite angles]
So, by AA-criterion of similarity, we have
                        increment FBE space tilde space increment FDA

rightwards double arrow                     space FB over FD equals FE over FA

rightwards double arrow                       FB over DF equals EF over FA

rightwards double arrow             DF cross times EF equals FB cross times FA.


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