Question

ABCD is a parellelogram, AD is produced to E so that DE = DC = AD and EC produced meets AB produced in F. Prove that BF = BC.

Answer 1

Given: In ||gm ABCD, AB is produced to E so that DE = DA and EC produced meets AB produced in F. To prove: BF = BC

Proof : In $\Delta ACE,$

O and D are the mid points of sides AC and AE

$\therefore$ DO || EC and DB || FC

$\Rightarrow BD||EF$

$\therefore AB=BF$

But AB = DC (Opposite sides of $|{|}^{gm}$)

$\therefore$ DC = BF

Now in $\Delta EDCand\Delta CBF,$

DC = BF (proved)

$\angle EDC=\angle CBF$

$(\because \angle EDC=\angle DAB$ corresponding angles)

$\angle DAB=\angle CBF$ (corresponding angles)

$\angle ECD=\angle CFB$ (corresponding angles)

$\therefore \Delta EDC\cong \Delta CBF$ (ASA criterion)

$\therefore DE=BC$ (c.p.c.t)

$\Rightarrow DC=BC$

$\Rightarrow AB=BC$

$\Rightarrow BF=BC$ ($\because$ AB = BF proved) \)

Hence proved.