Triangle is Convex Set

Theorem
The interior of a triangle embedded in $\R^2$ is a convex set.

Proof
Let $A_1, A_2, A_3 \in \R^2$ denote the vertices of the triangle.

For $i \in \left\{ {1, 2, 3}\right\}$, let $\mathcal L_i'$ and $\mathcal L_i''$ denote the rays that begin at $A_i$ and extend the sides adjacent to the angle $\angle A_i$.

Then, the opposite side of $\angle A_i$ becomes a line segment between $\mathcal L_i'$ and $\mathcal L_i''$.

From Sum of Angles of Triangle Equals Two Right Angles, it follows that each $\angle A_i$ is smaller than $\pi$ radians.

Let $S_i$ denote the set of points between the rays $\mathcal L_i'$ and $\mathcal L_i''$.

From Interior of Convex Angle is Convex Set, it follows that each $S_i$ is a convex set.

Let $P \in R^2$ be a point in the interior of the triangle.

Then $P$ lies between $\mathcal L_i'$ and $\mathcal L_i''$ for all $i \in \left\{ {1, 2, 3}\right\}$, so $\displaystyle P \in \bigcup_{i \mathop = 1}^3 S_i$.

If $Q \in R^2$ is a point not in the interior of the triangle, then $Q$ lies either outside or on the boundaries of the triangle.

So there is at least one $i_0 \in \left\{ {1, 2, 3}\right\}$ such that $Q \notin S_{i_0}$.

It follows that the interior of the triangle is equal to the intersection $\displaystyle \bigcup_{i \mathop = 1}^3 S_i$.

The result now follows from Intersection of Convex Sets is Convex Set.