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

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Two parallel lines AB and CD and a transversal EF intersect them at G and H respectively. GM, HM, GL and HL are the bisectors of the two pairs of interior angles.
To prove that GMHL is a rectangle
∵AB∥CD
∴∠AGH=∠DHG (Alternate interior angles)
∠1=∠2
(GM & HL are bisectors of ∠AGH and ∠DHG respectively)
⇒GM∥HL
(∠1 and ∠2 from a pair of alternate interior angles and are equal)
Similarly, GL∥MH
So, GMHL is a parallelogram.
∵AB∥CD
∴∠BGH+∠DHG=180
(Sum of interior angles on the same side of the transversal =180)
∠3+∠2=90
(GL & HL are bisectors of ∠BGH and ∠DHG respectively).
In ΔGLH,∠2+∠3+∠L=180
∠L=180-90
⇒∠L=180
Thus, in parallelogram GMHL, ∠L=90
Hence, GMHL is a rectangle.

Final Answer: If two parallel lines are intersected by a transversal, then bisectors of the interior angles from a rectangle.

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