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

          
Given:
AB and CD are two parallel lines intersected by a transversal EF at G and H respectively.
GM is the bisector of ∠ AGH, GL is the bisector of ∠ BGH, HL is the bisector of ∠ DHG, HM is the bisector of ∠ CHG.
To Prove: GMHL is a rectangle.
Proof:
AB || CD and EF intersects them.
     ∠ AGH = ∠ GHD [Alternate angles]
⇒ ∠ AGH = ∠ GHD
     ⇒ ∠ 1 = ∠ 2 …………………(i)
But these angles form a pair of equal alternate angles for lines GM and HL and transversal GH.
⇒ GM || HL
Similarly HM || GL.
⇒ GLHM is a parallelogram.
⇒ AB || CD and EF is the transversal.
∴ ∠ BGH + ∠ GHD = 180° [ The sum of the interior angles on the same side of the transversal is 180° ]
⇒ ∠ BGH + ∠ GHD = = 90°
             ⇒ ∠ 3 + ∠ 2 = 90° ……………………….(ii)
In Δ GHL, ∠ 3 + ∠ 2 + ∠ GLH = 180° [Sum of the three angles of a triangle is 180° ]
           ⇒ 90° + ∠ GLH = 180°
  ⇒ ∠ GLH = 180° - 90° = 90°
⇒ One angle of the parallelogram is a right angle.
⇒ Parallelogram GLHM is a rectangle.