Shortest distance between two non-intersecting differentiable curves is along their common normal I want to formally prove the following property:
The shortest distance between two differentiable non-intersecting curves is along their common normal.
I looked at the discussion on this on Quora, (Link: https://www.quora.com/How-can-I-show-that-the-shortest-distance-between-2-non-intersecting-curves-always-lies-along-their-common-normal), but the proofs were not formal there.
My approach:
Let the curves be denoted by f and g respectively. And let AB be a line segment with A lying on f and B lying on g. 
WLOG, take A as the origin of the coordinate axis, and the tangent to f at A to be the x-axis.
Let $B \equiv (b,g(b))$, $A = (0,0)$. Let the angle AB makes with the x-axis be $\theta \in (0,\pi)$.
Let 2 rays from A making an angle of $\theta + \epsilon$ and $\theta - \epsilon$ ($\epsilon > 0$) with the x-axis intersect $y = g(x)$ at $B_1$ and $B_2$ respectively.
I wish to show that $\theta \neq \frac{\pi}{2} \Leftrightarrow |AB| > \min\{|AB_1|, |AB_2| \}$.
Suppose $\theta \neq \frac{\pi}{2} $. Then let $B_1 \equiv (b_1,g(b_1))$ and $B \equiv (b_2,g(b_2))$.Then 
\begin{align}
g(b) &= b\tan(\theta)\\
g(b_1) &= b_1\tan(\theta + \epsilon)\\
g(b_2) &= b_2\tan(\theta - \epsilon) \\
\Rightarrow |AB| &= |b \sec(\theta) | \\
|AB_1| &= |b_1 \sec(\theta + \epsilon) | \\
|AB_2| &= |b_2 \sec(\theta - \epsilon) | 
\end{align}
I wanted to relate $b_1$ and $b_2$ to $b$ using the continuity of $g$. However I am not able to do so. How do I proceed from here? - or is there an easier formal approach?

Suppose $a(s),b(t)$ are two curves in $\mathbb R^2$ with parameter interval $(0,1).$ Assume $a'(s),b'(t)$ never vanish, otherwise normal vectors make no sense. Define
$$f(s,t)= |a(s)-b(t)|^2.$$
Suppose $f(s_0,t_0)>0$ is the minimum value of $f$ (hence $\sqrt {f(s_0,t_0)}$ is the distance between the two curves). Then $(s_0,t_0)$ is a critical point of $f.$ Thus
$$\frac{\partial f}{\partial s}(s_0,t_0) = 2a'(s_0)\cdot(a(s_0)-b(t_0)) = 0$$
and
$$\frac{\partial f}{\partial t}(s_0,t_0) = 2b'(t_0)\cdot(b(t_0)-a(s_0)) = 0.$$
(Here $\cdot$ denotes the dot product.) Thus both $a'(s_0),b'(t_0)$ are perpendicular to the vector $b(t_0)-a(s_0).$ Thus the vector $b(t_0)-a(s_0)$ is perpendicular to $a$ at $a(s_0),$ and is  perpendicular to $b$ at $b(t_0).$ This is the desired conclusion. 
