In Fig. POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

∠ROS = $\dfrac{1}{2}$ (∠QOS – ∠POS).

Since, OR ⊥ PQ

∴ ∠ROP = 90° and ∠ROQ = 90°

∴ ∠ROS = 90° - ∠POS …….(1)

⇒ ∠QOS = ∠QOR + ∠ROS

⇒ ∠QOS = 90° + ∠ROS

⇒ 90° = ∠QOS - ∠ROS …..(2)

Substituting value of 90° from equation (2) in equation (1), we get :

⇒ ∠ROS = (∠QOS - ∠ROS) - ∠POS

⇒ ∠ROS + ∠ROS = ∠QOS - ∠POS

⇒ 2(∠ROS) = ∠QOS - ∠POS

⇒ ∠ROS = $\dfrac{1}{2}$ (∠QOS - ∠POS)

Hence, proved ∠ROS = $\dfrac{1}{2}$ (∠QOS - ∠POS).