Question

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

Answer 1

The figure shown below shows two parallel lines AB and CD cutting by a transversal l;   X and Y are the points of intersection of l with AB and CD respectively. XP, XQ, YP and YQ are the angle bisectors of ∠AXY, ∠BXY, ∠CYX and ∠DYX respectively. AB∥CD and l is a transversal,So; ∠AXY = ∠DYX    (Pair of alternate angles)⇒12∠AXY = 12∠DYX⇒∠1 = ∠4             ⇒PX∥YQ ...(i)           {If a transversal intersects two lines in such a way that a pair of alternate interior angles are equal, then two lines are parallel}Also;  ∠BXY = ∠CYX      (Pair of alternate angles)⇒12∠BXY = 12∠CYX⇒∠2 = ∠3⇒PY∥XQ  ,...(ii)       {If a transversal intersects two lines in such a way that a pair of alternate interior angles are equal, then two lines are parallel}So from (i) and (ii) we get;PXQY is a parallelogram.        ...(iii)∠CYD = 180°⇒12∠CYD = 90°⇒12(∠CYX+∠DYX) = 90°⇒12∠CYX+12∠DYX = 90°⇒∠3+∠4 = 90°⇒∠PYQ = 90°Hence proved.   Hope it helps..            (I'm sorry it doesn't have proper spaces)  

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.