proof of when is a point inside a triangle

Let $𝐮\in {ℝ}^2$, $𝐯\in {ℝ}^2$and $\mathrm{𝟎}\in {ℝ}^2$. Let’s consider the convex hullof the set $T=\{𝐮,𝐯,\mathrm{𝟎}\}$.By definition, the convex hull of $T$, noted $coT$, is the smallestconvex set that contains $T$. Now, the triangle ${\mathrm{\Delta }}_T$ spannedby $T$ is convex and contains $T$. Then $coT\subseteq {\mathrm{\Delta }}_T$.Now, every convex $C$ set containing $T$ must satisfy that $t𝐮+\left(1-t\right)𝐯\in C$,$t𝐮\in C$ and $t𝐯\in C$ for $0\le t\le 1$ (atleast the convex combination of the points of $T$ are contained in$C$). This means that the boundary of ${\mathrm{\Delta }}_T$ is containedin $C$. But then every convex combination of points of $\partial {\mathrm{\Delta }}_T$must also be contained in $C$, meaning that ${\mathrm{\Delta }}_T\subseteq C$for every convex set containing $T$. In particular, ${\mathrm{\Delta }}_T\subseteq coT$.

Since the convex hull is exactly the set containing all convex combinationsof points of $T$,

$${\mathrm{\Delta }}_T=coT=\{𝐱\in {ℝ}^2:𝐱=\lambda 𝐮+\mu 𝐯+(1-\lambda -\mu )\mathrm{𝟎},0\le \lambda ,\mu ,\le 1,0\le 1-\lambda -\mu \le 1\}$$

we conclude that $𝐱\in {ℝ}^2$ is in the trianglespanned by $T$ if and only if $𝐱=\lambda 𝐮+\mu 𝐯$with $0\le \lambda ,\mu ,\le 1$ and $0\le \lambda +\mu \le 1$.