AB, CD are two parallel lines, and a transversal l intersects AB at X and CD at Y. Prove that the bisectors of the interior angles form a parallelogram, with all its angles right angles, i.e., it is a rectangle.

To Prove: Bisectors of the interior angles form a parallelogram.

Given: AB, CD are two parallel lines, and a transversal l intersects AB at X and CD at Y.

Concept Used:

ASA Theorem: If two angles and one side of a triangle is equal to two angles and one side of another triangle, the triangles are similar.

Diagram:

Proof:

Interior Angles: ∠AXY, ∠BXY, ∠XYC, and ∠XYD

Now, we know that,

∠AXY = ∠XYD [Alternate angles]

Bisectors of these angles,

Also,

∠BXY = ∠XYC [Alternate angles]

Now, from the figure, we can see that,

∠2 = ∠4 and ∠1 = ∠3

In ΔXSY and ΔXQY

∠2 = ∠4 and ∠1 = ∠3

XY is common

Therefore,

ΔXSY ≈ ΔXQY

Now as the triangles are similar,

∠S = ∠Q

∠1+ ∠4 = ∠2 + ∠3

As the opposite angles are equal,

XQSY is a parallelogram.

Also,

∠X + ∠S + Q + ∠Y = 360˚

4∠X = 360˚

∠X = 90˚

All angles are 90˚.

Hence, all angles are 90˚, and thus XSQY is a rectangle.