Question

If two parallel lines are intersected by a transversal, then prove that bisectors of the interior angles form a rectangle.

Answer 1

Let,

mn and xz be two parallel lines intersected by a transversal T.A and B are the points of intersection of T with mn and xz respectively.

AP, AQ, BP and BQ are the angle bisectors of ∠ mAB, ∠ BAn, ∠ ABx and ∠ zBA.
Here,

mn || xz and T is transversal.
∴ ∠ mAB = ∠ zBA (alternate angles)
⇒ 1/2 ∠ mAB = 1/2 ∠ zBA
⇒ ∠alpha = ∠ beta (∠ alpha = 1/2 ∠ mAB and ∠ beta = 1/2 ∠ zBA)

Therefore, PA || BQ ….....(1)

Also ∠ nAB = ∠ xBA (alternate angles)
⇒ 1/2 ∠nAB = 1/2 ∠ xBA
⇒ ∠ gamma = ∠ delta (∠ gamma = 1/2 ∠ nAB and ∠ delta = 1/2 ∠ xBA)
Therefore, PB || QA …....(2)

From (1) and (2), we get
PAQB is a parallelogram ....(3)

Now,

∠ xBz= 180°
⇒ 1/2 ∠ xBz = 180/2 = 90°
⇒ 1/2 (∠xBA + ∠ zBA) = 90°
⇒ 1/2 ∠ xBA + 1/2 ∠ zBA = 90°
⇒ ∠delta + ∠ beta = 90°
⇒ ∠ PBQ = 90° ...(4)

So, using (3) and (4), we conclude that PBQA is a rectangle.

QED,………