Diameter of a triangle If $T\subseteq \mathbb R^2$ is a generic plane triangle, I want to find its diameter
$$d=\sup\{\lvert| x-y \rvert|: x,y\in T\}$$
Intuitively I think that $d$ is the length of the longest edge of $T$. How can I formally prove this?
 A: Let the vertices be $a_1,a_2,a_3$. Then any $x,y \in T$ can be written as convex combinations of these, so let $x = \sum \lambda_i a_i$ and $y = \sum \mu_i a_i$ with $\sum \lambda_i = \sum \mu_i = 1$. Then by the triangle inequality
$$\newcommand{\norm}[1]{\lVert #1 \rVert}
\norm{x - y} \le \sum_{i, j} \lambda_i \mu_j \norm{a_i - a_j} \le \left(\sum \lambda_i \mu_j\right) \left(\max_{i, j}\ \norm{a_i - a_j}\right) = \max_{i, j}\ \norm{a_i - a_j}
$$
Thus, the diameter $\sup_{x, y} \norm{x - y} \le \max_{i, j} \norm{a_i - a_j}$. The reverse inequality holds by taking $x = a_i$, $y = a_j$.
So the same proof works for any convex polytope in $\mathbb{R}^n$ once you replace edge by "segment between two vertices".
A: Given any triangle $T$, in fact any bounded closed convex subset of $\mathbb{R}^2$.
The map
$$T^2 \ni (x,y) \mapsto |x-y|^2 \in \mathbb{R}$$
is a continuous function on $T^2$ bounded from above. Since $T^2$ is compact, the map achieves its maximum on some $(u, v) \in T^2$. i.e.
$$\sup \{\;|x-y|^2 : x, y \in T\;\} = | u - v |^2.$$
If either $u$ or $v$ is not an extremal point
of $T$, say $u = \frac12 ( u_1 + u_2 )$ where $u_1, u_2 \in T$, then by substituting $x_1$ by $u - v$ and $x_2$ by $\frac12 ( u_1 - u_2 )$ into 
parallelogram identity:
$$| x_1 + x_2 |^2 + |x_1 - x_2|^2 = 2 ( |x_1|^2  + |x_2|^2 )$$
one find at least one of $|u_1 - v|^2$ and $|u_2 - v|^2$ is greater than $|u-v|^2$. This contradicts with the role of $u, v$ that maximize $|u - v|^2$. 
As a result, the $u, v$ that maximize $|u-v|^2$ are both extremal points and
$$\sup\{\;|x-y|^2 : x,y \in T\;\} = \max\{\;|u-v|^2 : u, v \in T, \text{ both extremal}\;\}$$
A triangle has 3 extremal points, i.e. its 3 vertices, and hence its diameter is the length of its longest edge. 
A: [intuitive answer]
If $y$ is fixed and if $x$ is a point in interior, we draw a small ball $B(x,r)$. 
Then there exists a point $x'\in \partial B(x,r)$ such that 
$$ d(x',y) =d(x,y)+r $$
