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
Answers
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)
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.
