Interior of Convex Angle is Convex Set

Theorem
Let $\mathbf v, \mathbf w$ be two non-zero vectors in $\R^2$, and let $p$ be a point in $\R^2$.

Suppose that the angle between $\mathbf v$ and $\mathbf w$ is a convex angle.

Then the set


 * $U = \left\{ {p + st \mathbf v + \left({1-s}\right) t \mathbf w : s \in \left({0\,.\,.\,1}\right), t \in \R_{>0} }\right\}$

is a convex set.

Proof
Let $p_1 ,p_2 \in U$.

Then for $i \in \left\{ {1, 2}\right\}$, $p_i = p + s_i t_i \mathbf v + \left({1 - s_i}\right) t_i \mathbf w$ for some $s_i \in \left({0\,.\,.\,1}\right), t_i \in \R_{>0}$.


 * [[File:InteriorOfConvexAngle.png]]

WLOG assume that $t_1 \le t_2$.

Suppose that $q \in \R^2$ lies on the line segment joining $p_1$ and $p_2$, so:

As $t_1 \le t_2$, it follows that $r \in \R_{>0}$.

We have $\dfrac{ \left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}> 0$, and:

It follows that $\dfrac{ \left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r} \in \left({0\,.\,.\,1}\right)$.

Then $q \in U$.

By definition of convex set, it follows that $U$ is convex.