If two parallel lines intersected by a transversal; prove that the bisectors of the two pairs of interior angle encloses a rectangle.

Explanation

Given : AB||CD and l cuts them
Aslo EF, GF, EH and GH are bisectors of `angleAEG, angleEGC, angle EGD` respectively,
`i.e., angle1=angle2,angle3=angle4,angle5=angle6 and angle7 = angle8`
To Prove EFGH is a rrectangle
Proof : Since AB|| CD and l cuts them,
`:. angle1+angle2=angle5+angle6`
(all `angle` formed with AB, CD and l i.e., 3 necessary line)

`rArr angle2+angle2=angle+5+angle5 " " ( :. angle1 = angle2 and angle5 = angle6)`
`rArr 2 angle2=2 angle5`
`rArr angle2=angle5`
But these are the alternate angles formed with 3 necesaary lines EF, GH and l
`:. EF||GH" ".....(1)`
Similarly, we can prove that `FG||GH " ".....(2)`
`:.` From eqs. (1) and (2), we get
EFGH is a parallelogram `" "` (pairs of apposite sides are parallel)....(3)
Now, `angle1+angle2+angle3+angle4=180^(@) " "(L.P.A.)`
`rArr angle2+angle2+angle3+angle3=180^(@) ( :. angle1= angle2 and angle3 = angle4` , as EF and EH are angle bisectors)
`:. 2 angle2+2angle3=180^(@)`
`rArr angle2+angle3=90^(@)" ".....(4)`
So, from eqs. (3) and (4) we can say that
EFGH is a reactangle. `" "` ( `:.` parallelogram has one right agnle)