Bisection of Angle in Cartesian Plane/Corollary

Corollary to Bisection of Angle in Cartesian Plane

Let $\theta$ be the angular coordinate of a point $P$ in a polar coordinate plane.

Let $QOR$ be a straight line that bisects the angle $\theta$.

If $\theta$ is in quadrant I or quadrant II, then the angular coordinates of $Q$ and $R$ are in quadrant I and quadrant III.

If $\theta$ is in quadrant III or quadrant IV, then the angular coordinates of $Q$ and $R$ are in quadrant II and quadrant IV.

Proof

From Bisection of Angle in Cartesian Plane, the angular coordinates of $Q$ and $R$ are $\dfrac \theta 2$ and $\pi + \dfrac \theta 2$.

Without loss of generality, let $\angle Q = \dfrac \theta 2$ and $\angle R = \pi + \dfrac \theta 2$.

Let $\theta$ be in quadrant I or quadrant II.

Then $0 < \theta < \pi$.

Dividing each term in the inequality by $2$:

$0 < \dfrac \theta 2 < \dfrac \pi 2$

and so $Q$ lies in quadrant I.

Adding $\pi$ to each expression in the inequality:

$\pi < \pi + \dfrac \theta 2 < \dfrac {3 \pi} 2$

and so $R$ lies in quadrant I.

$\Box$

Let $\theta$ be in quadrant III or quadrant IV.

Then $\pi < \theta < 2 \pi$.

Dividing each term in the inequality by $2$:

$\dfrac \pi 2 < \dfrac \theta 2 < \pi$

and so $Q$ lies in quadrant II.

Adding $\pi$ to each expression in the inequality:

$\dfrac {3 \pi} 2 < \pi + \dfrac \theta 2 < 2 \pi$ and so $R$ lies in quadrant IV.

$\blacksquare$