Question

If two parallel lines are intersected by a transversal, then price that the quadrilateral formed by the biscetors of two pairs of interior angles is a rectangle.​

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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.

To Prove: GMHL is a rectangle.

Proof:

∵AB∥CD

∴∠AGH=∠DHG (Alternate interior angles)

⇒ 1/2 ∠AGH= 1/2 ∠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°

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

⇒ 1/2∠BGH + 1/2

∠DHG=90°

⇒∠3+∠2=90°...(3)

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

In ΔGLH,∠2+∠3+∠L=180 °

⇒90° +∠L=180 °

Using (3)

⇒∠L=180° −90 °

⇒∠L=90°

Thus, in parallelogram GMHL, ∠L=90°

Hence, GMHL is a rectangle.