Question

1) if two lines are pralled then Prove
that the bisectors of alternate interior
angels fromed by tranvarsal are prallel​

Answer 1

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.

Step-by-step explanation:

To Prove: GMHL is a rectangle.

Proof:

∵AB∥CD

∴∠AGH=∠DHG (Alternate interior angles)

⇒

2

1

∠AGH=

2

1

∠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=180

o

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

o

)

⇒

2

1

∠BGH+

2

1

∠DHG=90

o

⇒∠3+∠2=90

o

.....(3)

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

In ΔGLH,∠2+∠3+∠L=180

o

⇒90

o

+∠L=180

o

Using (3)

⇒∠L=180

o

−90

o

⇒∠L=90

o

Thus, in parallelogram GMHL, ∠L=90

o

Hence, GMHL is a rectangle.

solution

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