Accordingly to Pommerenke's Boundary Behaviour of conformal Maps, a John domain is a bounded simple connected domain in the plane such that there is a $M>0$ that satisfies that for every crosscut $[a,b]$ (i.e. a segment joining two points, $a$ and $b$, in the boundary of $G$) that $$diam H \leq M |a-b|,$$
where $diam$ is the usual euclidean diameter and $H$ is the component of $G\setminus[a,b]$ with smallest diameter. If $\partial G$ is smooth, then $G$ looks like a set in the plane not having outward-pointing cusps.
The problem: I need to show that every bounded convex set is a John domain. But I am very inexperienced with convex sets.
My attepmt
Let $G$ be a bounded convex set. I tried taking the tangent lines at $a$ and $b$, say $T_a$ and $T_b$ respectively. Assuming they are not parallel (?!), they meet at a point $v$, and with those lines and a segment parallel to $[a,b]$ I formed the triangle $AvB$ (see draw) such that $G$ is contained in its interior.
I need to show that $diam H/|a-b| \leq k'$ for some constant $k'$; but $diam H$ is less or equal than the longest side of the small triangle $avb$, lets call the length of this side $l$. Let $L$ be the length of the corresponding side of the big triangle $AvB$. Using this and the congruence of $avb$ with $AvB$ we have $$diam H/|a-b|\leq l/|a-b|=L/|A-B|.$$
If I were able to prove that the longest side of $AvB$ is no longer than a constant $K$ depending only on $G$ (our initial bounded convex set) and the shortest side is no shorter than a constant k, then I could write
$$diam H/|a-b|\leq L/|A-B| \leq K/k$$
with $K/k$ constant, i.e. not depending on the selected points $a$ and $b$, and we would be done... But I don't know how to argument the existence of $K$ and $k$ (if they exist) and still have the problem that all this work only when $T_a$ is not parallel to $T_b$...
Any suggestion, even using a completely different approach, would be highly appreciated.