Question

if two parallel lines are intersected by a transversal , prove that the bisector of interior angles on the same side of transversal intersect each other at right angles

Answer 1

Ello There!

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Question: If two parallel lines are intersected by a transversal , prove that the bisector of interior angles on the same side of transversal intersect each other at right angles.

(Based on NCERT Grade 9 Mathematics)

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Pre-Requisite Knowledge

• Alternate interior angles are equal.

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Answer

Given,

Interior angles on the same side of the transversal are bisected.

To Prove,

∠XSY = 90°

∠XQY = 90°

Proof,

∠BXy + ∠DYX = 180° (co-interior angles)

$\frac{1}{2}$∠BXY + $\frac{1}{2}$∠DYX = $\frac{1}{2}$180° (halves of equals are equal)

∠1 + ∠2 = 90°  (the angles are bisected) Let this be →1

Now,

In Δ XQY

∠1 + ∠2 + ∠XQY = 180° (A.S.P of triangles)

      90° + ∠XQY = 180° (From →1)

                ∠XQY = 180° - 90°

               ∠XQY = 90°       →2

Similarly we can prove that

               ∠XSY = 90°       →3

Therefore, Interior angles on the same side of transversal intersect each other at right angles.

(figure attached to refer)

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Hope It Helps!

Answer 2

Solutions:

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 at right angles.