Consider the points $O = (0,0)$, $A = (0,2)$, $B = (6,0)$ and $C = (6,3)$. Suppose there is a point $D = (h,0)$ on $OB$. Find $h$ so that $AD+DC$ is minimised.
We can define $$f(h)=AD+DC=\sqrt{4+h^2}+\sqrt{9+(6-h)^2}.$$ Then \begin{align} & f'(h) =0 \\ & \implies \frac{h}{\sqrt{h^2+4}}=\frac{6-h}{\sqrt{9+ (6-h)^2}} \\ & \implies h^2[9+(6-h)^2] =(6-h)^2[h^2+4] \\ & \implies 3h =2(6-h) \\ & \implies h =12/5. \end{align} Further, this gives the required minimum $f_{max}=\sqrt{61}.$
The question is how else this problem can be solved?
Let point $E(6;-3)$ be reflection of point $C(6;3)$ about x-axis.
then draw line segment $AE$ that is the minimum distance $AE$ through $D$.
Location of the point $D$ is the intersection of the line segment and x-axis.
$\textbf {Claim : $ADE$ is the minimum distance.}$
$\textbf {Proof : Take another point $F$ on the x-axis.}$
$\textbf {ADE < AF+FB (triangle inequality)}$
$\textbf {Since $\triangle DCE$ is an isosceles with $DC=DE$, $AD+DC=AE$}$
From point $A$ to $E$ the horizontal distance is $6$ and vertical distance is $5$.
$AE^2=6^2+5^2=61$, $AE=\sqrt{61}$ (Pytha)
$\triangle AOD \sim \triangle EBD$ (A.A.A. Similarity Theorem)
$\frac{AO}{OD}=\frac{EB}{BD}$
$\frac{2}{OD}=\frac{3}{6-OD}$
$3OD=12-2OD$
$OD=\frac{12}{5}$
Point $D(\frac{12}{5};0)$