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


 * $U = \set {p + s t \mathbf v + \paren {1 - s} t \mathbf w : s \in \openint 0 1, t \in \R_{>0} }$

is a convex set.

Proof
Let $p_1, p_2 \in U$.

Then for $i \in \set {1, 2}$, $p_i = p + s_i t_i \mathbf v + \paren {1 - s_i} t_i \mathbf w$ for some $s_i \in \openint 0 1, t_i \in \R_{>0}$.


 * [[File:InteriorOfConvexAngle.png]]

, 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:
 * $\dfrac {\paren {1 - s} s_1 t_1 + s s_2 t_2} r > 0$

and so:

It follows that:
 * $\dfrac {\paren {1 - s} s_1 t_1 + s s_2 t_2} r \in \openint 0 1$

Then $q \in U$.

It follows by definition that $U$ is convex.