Question

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

Answer 1

R.E.F image

Given : $ABCD$ is A parallelogram

To Prove : $BF=BC$

Proof : In $△DCE,DE=DC$ (given)

$\therefore \angle DCE=\angle DEC...\left(1\right)$

(Equal sides have equal is opposite to them)

since,

$AB\parallel CD,\angle DCE=\angle BFC...\left(2\right)$ (pair of corresponding $\angle S$)

Form (1) and (2)

$\angle DEC=\angle BFC$

In $△AEF,\angle AEF=\angle AFE$

$\therefore AF=AE,$

$\Rightarrow AB+BF=AD+DE$

$\Rightarrow BF=AD$ $[\because AB=CD=DE]$

$\Rightarrow BF=BC$ $[\because AD=BC]$ Hence proved.