Question

In Figure ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F, then AF/AB

Answer 1

Answer:

We know,

E is midpoint of BC.

∴BC = CE

Let consider triangles ΔCDE and ΔBEF

So,

∠CED = ∠BEF (vertically opp. angle)

∠DCE = ∠FBE (alternate angles)

So.....ΔCDE ≈  ΔBFE

Then CD = BF (CPCT)

∵CD = AB

∴AB = BF

→AF = AB + BF

→AF = AB + AB

→AF = 2AB        (1)

now, AF/AB= 2AB/AB= 2  (FROM 1)

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

ANGLE BEF = DEC ( VERTICALLY OPPOSITE ANGLES)

ANGLE EBF= DCE ( DC IS PARALLEL TO AF AND ALTERNATE INTERIOR ANGLES). *

SO BY ASA CONGRUENCE RULE TRIANGLE DCE AND BEF ARE CONGRUENT.

BY C.P.C.T BF = DC. - 1

DC = AB ( ABCD IS A PARALLELOGRAM) - 2

SO AF = AB PLUS BF.

SO FROM 1 AND 2 WE GET AF = 2AB