Two parallel lines are intersected by a transversal. Then, prove that the bisectors of two pairs of interior angles enclose a rectangle.

Given: Two parallel lines AB and CD are intersected by a transversal EF in points G and H respectively. The bisectors of two pairs of interior angles intersect in L and M.
To Prove: GLHM is a rectangle.
Proof: ∵ AB || CD and a transversal EF intersects them
 
                          | Alternate interior angles

                             | Halves of equals are equal

But these form a pair of equal alternate interior angles
∴ GM || HL    ...(1)
Similarly, we can show that
HM || GL    ...(2)
In view of (1) and (2),
GLHM is a parallelogram
| A quadrilateral is a parallelogram if its both the pairs of opposite sides are parallel

Now, since the sum the consecutive interior angles on the same side of a transversal is 180°

                             | Halves of equals are equal

⇒    ∠3 + ∠2 = 90°    ...(3)
In ∆GHL,
∠3 + ∠2 + ∠GLH = 180°
| Angle sum property of a triangle
⇒    90° + ∠GLH = 180°
| From (3)
∠GLH = 90°
⇒ GLHM is a rectangle
| A parallelogram with one of its angles of measure 90° is a rectangle.