Question

In figure 5.13 ABCD is a parallelogram.Point E is on the ray AB such that BE=AB then prove that line ED bisects seg BC at point F.

Answer 1

AB=DC
AB=BE
THEREFORE,BE=DC
In triangle DCF and EBF
ANGLE DCF= ANGLE EBF(corresponding angles)
angle CFD= ANGLE BFE( vertically opposite angle)
DC=BE(proved above)
∆ DCF =~∆EBF
therefore,bf=cf(cpct)
hence,ED BISECT BC

Answer 2

Given:

ABCD is a parallelogram .

Point E is on the ray AB such that

BE = AB .

To prove : BF = FC

Proof :

i ) AB = DC [ opposite sides ]

AB = BE ( given )

DC = BE --- ( 1 )

ii ) In ∆DCF and ∆EBF

<FDC = <FEB [ alternate angles ]

DC = BE ( S ) [ from ( 1 ) ]

<DFC = <EFB [ veritically opposite angles ]

Therefore ,

∆DCF is congruent to ∆EBF

[ ASA congruence Rule ]

CF = FB [ CPCT ]

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