Let’s prove that internal and external bisectors of an angle are perpendicular to each other.

Let us first draw a figure to support our proof.

Proof:

In the above figure PQ is a straight line. O is the point of intersection of line with ray RO, TO and SO.

Let ∠POS = x and let ∠QOS = y

Here ∠POS is an internal angle and ∠QOS is an external angle.

Since PQ is a straight line,

x + y = 180°

x = 180 – y ……. (i)

Let OR be the internal angle bisector.

So ∠POR = ∠SOR = 1/2 ∠POS [as angle bisector divides the given angle in two halves]

In the similar manner OT is the external angle bisector.

So, we can write ∠SOT = ∠TOQ = 1/2 ∠SOQ

Now when we combine, we get a common angle bisector ∠ROT.

∠ROT = ∠SOR + ∠SOT

∠ROT = 1/2 ∠POS + 1/2 ∠QOS

= 1/2 × (∠POS + ∠QOS)

= 1/2 × (x + y)

= 1/2 × (180 – y + y) ………. (From equation (i))

= 1/2 × 180

∠ROT = 90°

Since the angle is 90°, it is proved that internal and external bisectors of an angle are perpendicular to each other.