Minimal value of $\sqrt{(x-a)^2 + b^2} +\sqrt{(x-c)^2 + d^2}$ without derivatives and without distance formula Let  $f(x) = \sqrt{(x-a)^2 + b^2} +\sqrt{(x-c)^2 + d^2}$, where all coefficients are real. 
It can be shown using a distance formula that the minimal value of $f(x)$ is 
$D = \sqrt{(a-c)^2+(|b|+|d|)^2}$.
Show that result without derivatives and without a distance formula. At what value of $x$ does the minimum of $f(x)$ occur?
Hint: this is a generalization of this question.

Applying Fermat's Principle on reflection:
Assume $a<c$,
\begin{align*}
  T(x) &= \sqrt{(x-a)^2+b^2}+\sqrt{(x-c)^2+d^2} \\
  T'(x) &=\frac{x-a}{\sqrt{(x-a)^2+b^2}}-\frac{c-x}{\sqrt{(x-c)^2+d^2}} \\
  0 &= \sin i-\sin r \\
  i &= r \\
\end{align*}

If you've accepted the law of reflection, then
\begin{align*}
    \frac{x-a}{b} &= \frac{c-x}{d} \\
    x &= \frac{ad+bc}{b+d}
  \end{align*}
where $b,d>0$

See another answer for the case of refraction.
