Interior of Convex Angle is Convex Set

Theorem
Let $\mathbf v_1, \mathbf v_2$ be two non-zero vectors in $\R^2$, and let $P$ be a point in $\R^2$.

For $i \in \left\{ {1, 2}\right\}$, denote the ray that starts at $P$ and extends to infinity parallel with $v_i$ as:


 * $\mathcal L_i = \left\{ {Q \in \R^2 : Q = P + t \mathbf v_i, \ t \in \R_{\ge 0} }\right\}$

Suppose that the angle between $\mathbf v_1$ and $\mathbf v_2$ is a convex angle.

Then the set of points between the rays $\mathcal L_1$ and $\mathcal L_2$ is convex.

Proof
Let $T_1, T_2 \in \R^2$ lie between $\mathcal L_1$ and $\mathcal L_2$.

Then $T_i$ lies on the line segment between $P + t_i \mathbf v_1$ and $P + t_i \mathbf v_2$ for some $t_i \in \R_{>0}$ for $i \in \left\{ {1, 2}\right\}$:

WLOG assume that $t_1 \le t_2$.

If $T \in \R^2$ lies on the line segment between $T_1$ and $T_2$, then:

As $t_2 \ge t_1 > 0$ and $s > 0$, it follows that $t_0 \in \R_{> 0}$.

As $r_1, r_2 \in \left({0\,.\,.\,1}\right)$, it follows that $s_0 \in \left({0\,.\,.\,1}\right)$.

Hence, $T$ lies on the line segment between $t_0 \mathbf v_1 \in \mathcal L_1$ and $t_0 \mathbf v_2 \in \mathcal L_2$.

By definition of convex set, the result follows.