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Answer:

Given: 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: GMHL is a rectangle.

Proof:

∵AB∥CD

∴∠AGH=∠DHG (Alternate interior angles)

⇒21∠AGH=21∠DHG

⇒∠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=180o

(Sum of interior angles on the same side of the transversal =180o)

⇒21∠BGH+21∠DHG=90o

⇒∠3+∠2=90o .....(3)

(GL & HL are bisectors of ∠BGH and ∠DHG respectively).

In ΔGLH,∠2+∠3+∠L=180o

⇒90o+∠L=180o Using (3)

⇒∠L=180o−90o

⇒∠L=90o

Thus, in parallelogram GMHL, ∠L=90o

Hence, GMHL is a rectangle.


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