Given, OD ⊥ OE, OD and OE are the bisectors of ∠AOC and ∠BOC.
To show points A, O and B are collinear, i.e., AOB is a straight line.
Since OD and OE bisect angles ∠AOC and ∠BOC respectively,. . . . . . . .(i)
∴ ∠AOC=2∠DOC.......(i)
and ∠COB=2∠COE..........(ii)
On adding equations (i) and (ii), we get,
∠AOC+∠COB=2∠DOC+2∠COE
⇒ ∠AOC+∠COB=2(∠DOC+∠COE)
⇒ ∠AOC+∠COB=2∠DOE
⇒ ∠AOC+∠COB=2×90∘
⇒ ∠AOC+∠COB=180∘
∴ ∠AOB=180∘
∠AOC and ∠COB are forming linear pair.
Therefore, AOB is a straight line.
Hence, points A, O and B are collinear.