In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ ROS = $\frac{1}{2}$ (∠ QOS – ∠ POS).

Given: OR is perpendicular to PQ. ∠ROQ = ∠ROP = 90°.

To prove: ∠ROS = 1/2(∠QOS - ∠POS).

Let ∠ROS = a, ∠POS = b and ∠SOQ = c.

To prove that: a = 1/2(c - b).

Since ∠ROQ = ∠ROP = 90°,

We can say, ∠POS + ∠SOR = ∠POR

b + a = 90°.….(1)

Line PQ is intersected by ray OS.

Hence ∠POS + ∠SOQ = b + c = 180°

b + c = 180°….. (2)

From equation (1), we get a + b = 90°

Multiplying by 2 on both sides we get,

2(a + b) = 2 × 90°

2(a + b) = 180°…..(3)

Comparing equations (3) and (2),

2(a + b) = b + c

2a + 2b = b + c

2a = b + c - 2b

2a = c - b

a = 1/2 (c - b)

∴ ∠ROS = 1/2 (∠QOS - ∠POS)