Two straight lines intersect each other at a point and thus four angles are formed. Let’s prove that the bisectors of these angles are two perpendicular straight lines.

Let us first draw a diagram for the above question.

In the above figure,

PQ and SR are two straight lines.

O is the common point.

AO is the angle bisector of ∠POS.

NO is the angle bisector of ∠SOQ

BO is the angle bisector of ∠QOR

MO is the angle bisector of ∠POR.

∠POR = ∠SOQ [vertically opposite angles are always equal]

Similarly, ∠POS = ∠QOR [vertically opposite angles are always equal]

Let ∠POR = ∠SOQ = x and Let ∠POS = ∠QOR = y.

OR is perpendicular to line PQ.

So ∠POR + ∠QOR = 180° [as they form linear pair that is angles in a straight line always sum up to 180°]

∠MOR = ∠MOP = 1/2 ∠POR [as MO is the angle bisector]

Similarly, ∠BOR = ∠QOB = 1/2 ∠QOR [as BO is the angle bisector]

∠MOB = ∠MOR + ∠BOR

= 1/2 ∠POR + 1/2 ∠QOR

= 1/2 × (∠POR + ∠QOR)

= 1/2 × 180

∠MOB = 90°

So ∠SOQ + ∠QOR = 180° [as they form linear pair that is angles in a straight line always sum up to 180°]

∠NOS = ∠NOQ = 1/2 ∠SOQ [as NO is the angle bisector]

Similarly, ∠QOB = ∠BOR = 1/2 ∠QOR [as BO is the angle bisector]

∠NOB = ∠NOQ + ∠QOB

= 1/2 ∠SOQ + 1/2 ∠QOR

= 1/2 × (∠SOQ + ∠QOR)

= 1/2 × 180

∠NOB = 90°

Since the angle is 90°, it is proved that the bisectors of these angles are two perpendicular straight lines.