# 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 $s_i \in \left({0\,.\,.\,1}\right) , t_i \in \R_{>0}$.

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:

 $\displaystyle q$ $=$ $\displaystyle p + s_1 t_1 \mathbf v + \left({1 - s_1}\right) t_1 \mathbf w + s \left({ p + s_2 t_2 \mathbf v + \left({1 - s_2}\right) t_2 \mathbf w - p - s_1 t_1 \mathbf v - \left({1 - s_1}\right) t_1 \mathbf w }\right)$ for some $s \in \left({0\,.\,.\,1}\right)$ $\displaystyle$ $=$ $\displaystyle p + \left({ \left({1 - s}\right) s_1 t_1 + s s_2 t_2}\right) \mathbf v + \left({ \left({1 - s}\right) \left({1 - s_1}\right) t_1 + s \left({1 - s_2}\right) t_2}\right) \mathbf w$ $\displaystyle$ $=$ $\displaystyle p + \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}r \mathbf v + \dfrac{t_1 + st_2 - st_1 - \left({1 - s}\right) s_1 t_1 - s s_2 t_2}{r} r \mathbf w$ where $r = t_1 + s \left({t_2 - t_1}\right)$ $\displaystyle$ $=$ $\displaystyle p + \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}r \mathbf v + \left({ 1 - \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r} }\right) r \mathbf w$

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:

 $\displaystyle 1 - \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}$ $=$ $\displaystyle \dfrac{ \left({1 - s}\right) \left({1 - s_1}\right) t_1 + s \left({1 - s_2}\right) t_2}{r}$ $\displaystyle$ $>$ $\displaystyle 0$

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.

$\blacksquare$