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.

1489 Views

Triangles

Hope you found this question and answer to be good. Find many more questions on Triangles with answers for your assignments and practice.

Mathematics

Browse through more topics from Mathematics for questions and snapshot.