Question

In the figure, ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F, prove that AF = 2AB.

Answer 1

Given: In ||gm ABCD, E a midpoint of BC. DE is joined and produced to meet AB produced at F.

To prove :AF = 2AB

Proof : In $\Delta CDEand\Delta EBF$

$\angle DEC=\angle BEF$ (vertically opposite angles)

CE =EB (E is mid point of BC)

$\angle DCE=\angle EBF$ (alternate angles)

$\therefore \Delta CDE\cong \Delta BFE$ (SAS Axiom)

$\therefore$ DC = BF (c.p.c.t.)

But AB = DC (opposite sides of a || gm)

$\therefore$ AB = BF

Now, AF = AB+BF= AB+AB

= 2AB

Hence AF = 2AB