if two parallel lines are intersected by a transversal,then prove that bisectors of the interior angles form a rectangle.

Given-,EG , FG, EH and FH are bisectors of the interior angles ∠AEH, ∠CFE, ∠BEF and ∠EFD respectively..

To prove :- EFGH is a rectangle.

PROOF:- AB || CD and a transversal t cuts them at E anf F respectively .

:. ∠AEF=∠EFD [ alternate interior ∠s]

= 1/2 ∠AEF=1/2∠EFD = ∠GEF=∠EFH

but , these are alternate interior angles formed when the transversal EF cuts EG and FH

:. EG || FH. simillarly , EH|| FG

:. EGFH is a parallelogram.

Now,  ray EF stands on AB.

:. ∠AEF + ∠BEF = 180° (linear pair)

= 1/2∠AEF + 1/2∠BEF= 90° =  ∠GEF + ∠ HEF= 90°

= ∠GEH =  90° [ ∠GEF + ∠ HEF= ∠GEH] 

Thus, EFGH is a parallelogram , one of whose angles is 90°

:. EFGH is a rectangle..