If two parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of interior angles enclose a rectangle.
Answers
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.
Answer:
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