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..ans. .. .