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.
$\blacksquare$