In Fig. 6.9, OD is the bisector of ∠AOC, OE is the bisector of ∠BOC and

From the figure, it is given to us

OD⊥OE

⇒ ∠DOE = 90° - - - - (i)

OD is the bisector of ∠AOC.

OE is the bisector of ∠BOC.

We have to show that the points A, O and B are collinear, i.e.,

To show that AOB is a straight line. - - - - (ii)

OD is the bisector of ∠AOC

⇒ ∠AOD = ∠COD

⇒ ∠AOC = 2 × ∠COD - - - - (ii)
Similarly, OE is the bisector of ∠BOC

⇒ ∠BOE = ∠COE

⇒ ∠BOC = 2 × ∠COE - - - - (iii)

Adding equation (ii) and equation (iii),

∠AOC + ∠BOC = 2 × ∠COD + 2 × ∠COE

⇒ ∠AOC + ∠BOC = 2 × (∠COD + ∠COE)

From the figure, we can see that

∠COD + ∠COE = ∠DOE - - - - (iv)

Substituting equation (iv) in the above equation,

∠AOC + ∠BOC = 2 × (∠COD + ∠COE)

⇒ ∠AOC + ∠BOC = 2 × ∠DOE

⇒ ∠AOC + ∠BOC = 2 × 90° (From equation (i), we have ∠DOE = 90°)

⇒ ∠AOC + ∠BOC = 180°

⇒ ∠AOB = 180° (From the figure, ∠AOC + ∠BOC = ∠AOB) - - - - (iv)

From (iv), we can say that

∠AOC and ∠BOC are forming linear pair of angles.

Since, ∠AOC and ∠BOC are two adjacent angles and their sum is 180°, where OC is a ray standing on AOB,

It is true that A, O and B are collinear, thus making AOB a straight line.