Question

In figure OP bisects angle BOC and OQ bisects angle AOC show that angle POQ = 90 degree

Answer 1

Solution:-
Since OP bisects ∠ BOC.
∴ ∠ BOC = 2(∠ POC) ...(1)
Again, OQ bisects ∠ AOC.
∴ ∠ AOC = 2(∠ QOC) ... (2)
Since ray OC stands on line AB.
∴ ∠ AOC + ∠ BOC = 180°
⇒ 2(∠ QOC) + 2(∠ POC) = 180°  [Using (1) and (2)]
⇒ 2(∠ POC + ∠ QOC) = 180°
⇒ ∠ POC + ∠ QOC = 90°
⇒ ∠ POQ = 90°
Hence ∠ POQ = 90°  Proved.

Answer 2

Ray OP bisects ∠ BOC 
... ∠ COP  = ∠ POB 
... ∠ COP  =  1/2 ∠ BOC           ------------(1)
Ray OQ bisects ∠ AOC
... ∠ AOQ  = ∠ QOC
      ∠ QOC = 1/2 ∠ AOC             -----------(2)
Adding (1) and (2) 
... ∠COP + ∠ QOC = 1/2 [ ∠BOC + ∠ AOC ]    (∠AOC and ∠ BOC form a linear pair.)
                        = 1/2 x 180°                      (Linear Pair Axiom)
                ∠QOP =  90°
            ... ∠POQ is a right angle.