Triangle is Convex Set

Theorem

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

Proof

Denote the triangle as $\triangle$, and the interior of the boundary of $\triangle$ as $\operatorname{Int} \left({\triangle}\right)$.

From Boundary of Polygon is Jordan Curve, it follows that the boundary of $\triangle$ is equal to the image of a Jordan curve, so $\operatorname{Int} \left({\triangle}\right)$ is well-defined.

Denote the vertices of $\triangle$ as $A_1, A_2, A_3$.

For $i \in \left\{ {1, 2, 3}\right\}$, put $j = i \bmod 3 + 1$, $k = \left({i+1}\right) \bmod 3 + 1$, and:

$U_i = \left\{ {A_i + st \left({A_j - A_i}\right) + \left({1-s}\right) t \left({A_k - A_i}\right) : s \in \left({0\,.\,.\,1}\right) , t \in \R_{>0} }\right\}$

Suppose that the angle $\angle A_i$ between is $A_j - A_i$ and $A_k - A_i$ is non-convex.

As $\angle A_i$ is an internal angle in $\triangle$, it follows from definition of polygon that $\angle A_i$ cannot be zero or straight.

Then $\angle A_i$ is larger than a straight angle, which is impossible by Sum of Angles of Triangle Equals Two Right Angles.

It follows that $\angle A_i$ is convex.

From Characterization of Interior of Triangle, it follows that:

$\displaystyle \operatorname{Int} \left({\triangle}\right) = \bigcap_{i \mathop = 1}^3 U_i$

From Interior of Convex Angle is Convex Set, it follows for $i \in \left\{ {1, 2, 3}\right\}$ that $U_i$ is a convex set.

The result now follows from Intersection of Convex Sets is Convex Set (Vector Spaces).

$\blacksquare$