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# In Fig. 6.17, $ \mathrm{POQ} $ is a line. Ray $ \mathrm{OR} $ is perpendicular to line $ \mathrm{PQ} $. OS is another ray lying between rays $ O P $ and OR. Prove that

$ \angle \mathrm{ROS}=\frac{1}{2}(\angle \mathrm{QOS}-\angle \mathrm{POS}) $

"

Given:

$POQ$ is a line, Ray $OR$ is perpendicular to line $PQ$ and $OS$ is another ray lying between rays $OP$ and $OR$.

To do:

We have to prove that $\angle ROS = \frac{1}{2}(\angle QOS - \angle POS)$.

Solution:

Ray $OR \perp POQ$.

This implies,

$\angle POR = 90^o$

$\angle POS + \angle ROS = 90^o$.....…(i)

$\angle ROS = 90^o - \angle POS$

$\angle POS + \angle QOS = 180^o$ (Linear pair)

$= 2(∠POS + ∠ROS)$ [From (i)]

$\angle POS + \angle QOS = 2\angle ROS + 2\angle POS$

$2\angle ROS = \angle POS + \angle QOS - 2\angle POS$

$2\angle ROS =\angle QOS - \angle POS$

$\angle ROS = \frac{1}{2}(\angle QOS - \angle POS)$

Hence proved.

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