Question

ABCD is a parallelogram. AD is produced to E such that DE =DC. EC is produced to intersect AB produced in F.Prove that BF =BC.

Answer 1

Given : ABCD is a parallelogram. AD is produced to E such that DE =DC. EC is produced to intersect AB produced in F.
To prove : BF =BC
In $\Delta DCE,$
DE =DC (Given)
$\therefore \angle DCE=\angle DEC....\left(1\right)$ (In a triangle, equal sides have equal angles opposite to them)
AB||CD (AB lies on AF)
AF||CD and EF is the transversal,
$\therefore \angle DCE=\angle BFC...\left(2\right)$ (Pair of corresponding angles)
From (1) and (2), we get
$\angle DEC=\angle BFC$
In $\Delta AFE,$
$\angle AFE=\angle AEF(\angle DEC=\angle BFC)$
$\therefore AE=AF$ (In a triangle, equal angles have equal sides opposite to them)
$\Rightarrow AD+DE=AB+BF$
$\Rightarrow BC+AB=AB+BF$ (AD=BC, DE =CD and CD =AB $\Rightarrow AB=DE)$
$\Rightarrow BC=BF$