In the above figure #ABCD# is a parallelogram having #AB=3cm and AD=8cm#.
#AE#, the bisector of #/_BAD=2beta# intersects the side #BC=8cm# at #E#
Similarly #DF#, the bisector of #/_ADC=2alpha# intersects the side #BC=8cm# at #F#
We are to find out the lengths of #BE,EFandFC#
As #AE#, the bisector of #/_BAD=2beta# we have #/_DAE=BAE=beta#
Again #AD"||"BC# and #AE# is the intercept .
So #/_DAB="alternate"/_AEB=beta#
#=>"In " DeltaABE,/_BAE=BEA=beta#
So #BE=AB=3cm#
Similarly we can show that #DC=CF=3cm#
Hence residual #EF=BC-CF-BE=8-3-3=2cm#
So we have #BE=CF=3cmandEF=2cm#
