Defining a triangle with vertices $a$, $b$, $c$ by $t_1a+t_2b+t_3c$, where $0\leq t_1, t_2, t_3\leq 1$ and $1=t_1+t_2+t_3$. Why does this work? For $a$, $b$, $c$ in $\mathbb{R}^2$, I know that the set
$$A=t_1a+t_2b+t_3c$$
with $0\leq t_1,t_2,t_3\leq 1$ and $1=t_1+t_2+t_3$
is a triangle with vertices $a$, $b$, $c$.
Could someone explain to me how we came up with those parameters geometrically? How could you explain it and why do they work?
Thank you in advance.
 A: You forgot to add that $a,b,c\in\Bbb R^2$. Clearly, $a,b,c\in A$; for intance, $a=1\times a+0\times b+0\times c$. Furthermor, the line segment that goies from $a$ to $b$ is a subset of $A$, since any $p$ from that line segment can be written as $(1-t)a+tb$, for some $t\in[0,1]$, and therefore $p=(1-t)a+tb+0\times c$. For the same reason, $A$ contains the line segments joining $a$ and $c$ and joining $b$ and $c$.
Let $p$ be an element of $A$. If $t_1,t_2,t_3\in[0,1]$ are such that $t_1+t_2+t_3=0$, and if $p=t_1a+t_2+t_3c$, then\begin{align}p&=t_1a+t_2b+\bigl(1-(t_1+t_2)\bigr)c\\&=(t_1+t_2)\left(\frac{t_1}{t_1+t_2}a+\frac{t_2}{t_1+t_2}b\right)+\bigl(1-(t_1+t_2)\bigr)c.\end{align}Since$$\frac{t_1}{t_1+t_2}+\frac{t_2}{t_1+t_2}=1\quad\text{and}\quad\frac{t_1}{t_1+t_2},\frac{t_2}{t_1+t_2}\geqslant0,$$the point$$\frac{t_1}{t_1+t_2}a+\frac{t_2}{t_1+t_2}b\tag1$$belongs to the line segment joining $a$ to $b$, and therefore $p$ belongs the line segment joining $(1)$ to $c$; so, $p$ belongs to the triangle.
This works in the opposite direction too: if $p$ belongs to the trangle, find the point $q$ of the line segment joining $a$ to $b$ such that $q$ $p$ and $c$ are collinear. Then $q=(1-t)a+tb$, for some $t\in[0,1]$, and $p=(1-t')q+t'c$ for some $t'\in[0,1]$. So\begin{align}p&=(1-t')\bigl((1-t)a+tb\bigr)+t'c\\&=\bigl((1-t')(1-t)\bigr)a+(1-t')tb+t'c\\&\in A.\end{align}
