Question

If two parallel lines are cut by a transverssal, prove that the bisectors of the interior angles on the same side of the transverssal intersect each other at right angles.

Answer 1

Answer:

We know that the sum of interior angles on the same side of the transversal is 180°.

Hence, ∠BMN + ∠DNM = 180°

=> 1/2∠BMN + 1/2∠DNM = 90°

=> ∠PMN + ∠PNM = 90°

=> ∠1 + ∠2 = 90° ............. (i)

In △PMN, we have

∠1 + ∠2 + ∠3 = 180° ......... (ii)

From (i) and (ii), we have

90° + ∠3 = 180°

=> ∠3 = 90°

=> PM and PN intersect

Answer 2

Answer:

Let the angle at which the transversal intersects the lines be θ

\So, ∠BAD=θ,∠EBA=180−θ

OB bisects the ∠EBA and OA bisects the ∠BAD.

consider △AOB,∠BAO=2θ

∠OBA=21(180−θ)=90−2θ

∠BAO+∠OBA+∠BOA=180°

⟹90−2θ+2θ+∠AOB=180°

⟹∠AOB=90°

∴  the bisectors of internal angles on the same side of the transversal intersects at right angles.