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
Draw a parallelogram ABCD with AC and BD intersecting at O.
Produce AD to E such that DE = DC.
Join EC and produce it to meet AB produced at F.
In $\triangle\text{DCE},$
$\therefore\angle\text{DCE}=\angle\text{DEC}\dots(1)$ (In a triangle, equal sides have equal angles opposite to them)
AB || CD (Opposite sides of the parallelogram are parallel)
$\therefore$ AF || CD (AB lies on AF)
AF || CD and EF is the transversal,
$\therefore\angle\text{DCE}=\angle\text{BFC}\dots(2)$ (Pair of corresponding angles)
From (1) and (2), we get
$\angle\text{DCE}=\angle\text{BFC}$
In $\triangle\text{AFE},$
$\angle\text{AFE}=\angle\text{AEF}$ $(\angle\text{DEC}=\angle\text{BFC})$
$\therefore$ AE = AF (In a triangle, equal angles have equal sides opposite to them)
⇒ AD + DE = AB + BF
⇒ BC + AB = AB + BF $(\because$ AD = BC, DE = CD AND CD = AB, AB = DE$)$
⇒ BC = BF
Produce AD to E such that DE = DC.
Join EC and produce it to meet AB produced at F.
In $\triangle\text{DCE},$
$\therefore\angle\text{DCE}=\angle\text{DEC}\dots(1)$ (In a triangle, equal sides have equal angles opposite to them)
AB || CD (Opposite sides of the parallelogram are parallel)
$\therefore$ AF || CD (AB lies on AF)
AF || CD and EF is the transversal,
$\therefore\angle\text{DCE}=\angle\text{BFC}\dots(2)$ (Pair of corresponding angles)
From (1) and (2), we get
$\angle\text{DCE}=\angle\text{BFC}$
In $\triangle\text{AFE},$
$\angle\text{AFE}=\angle\text{AEF}$ $(\angle\text{DEC}=\angle\text{BFC})$
$\therefore$ AE = AF (In a triangle, equal angles have equal sides opposite to them)
⇒ AD + DE = AB + BF
⇒ BC + AB = AB + BF $(\because$ AD = BC, DE = CD AND CD = AB, AB = DE$)$
⇒ BC = BF
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