Question

In the figure, if AOB is a line, OP bisects BOC and OQ bisects < AOC, show that Z POQ is a right angle.​

Answer 1

✎ Given:

Given, AOB is a line wherein OP bisects ∠BOC and OQ bisects ∠AOC.

✎ To prove:

We have to prove that ∠POQ is 90°.

✎ Proof:

OP bisects ∠BOC so :

➠ ∠BOP = ∠POC . . . . . ( eqⁿ . 1)

OQ bisects ∠AOC so :

➠ ∠AOQ = ∠QOC . . . . . ( eqⁿ . 2)

From the figure :

➠ ∠POC = ∠POC + ∠QOC . . . . . (eqⁿ . 3)

We know that, ∠AOB = 180° because it is a straight line.

➠ ∠BOP + ∠POC + ∠QOC + ∠AOQ = 180°

From eqⁿ . 1 and 2, we get :

➠ 2∠POC + 2∠QOC = 180°

Removing 2 as the common term :

➠ 2 (∠POC + ∠QOC) = 180°

➠ ∠POC + ∠QOC = 180°/2

➠ ∠POC + ∠QOC = 90°

From eqⁿ . 3, we get :

➠ ∠POC = 90°

Henceforth, proved!