Question

if a transversal intersects two Parallel Lines prove that the bisectors of interior angles on the same side of transversal make a right angle at the point of intersection

Answer 1

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.

Answer 2

Given

• AB || CD and ED cuts it at H and G and bisector of ∠BHG and ∠DGH meet at I.

To prove

$∠HIG = \: {90}^{o}$

AB || CD and EF is transversal

$∠BHG + ∠DGH = {180}^{o}$

$\frac{1}{2} ∠BHG + \frac{1}{2} ∠DGH = \frac{ {180}^{o} }{2}$

$∠HGI + ∠GHI = {90}^{o}$

In triangle HIG

$∠HIG + (∠HGI + ∠GHI ) = {180}^{o}$

$∠HIG + {90}^{o} = {180}^{o}$

$∠HIG = {180}^{o} - {90}^{o}$

$∠HIG = {90}^{o}$

So bisector of interior angle on the same side of transversal make right angle at the point of intersection.

Hence proved.

Note: Refer the attached picture for the diagram.