Minimize $\| ACE \|$ by geometrical means I have the following figure 

Where $AB=10$m, $BD=12$m and $DE=12$m. The point C can slide
along the segment BD. Now the problem is to minimize the distance from A to D 
going along the dashed line. The problem can be solved using simple analysis and differentiation. Let $BC=x$ then 
$$\|ACE\| = f(x) = \sqrt{10^2-x^2}+\sqrt{(12-x)^2+12^2}$$. in order to find a minima which we know exists one would have to solve $f'(x)=0$ to obtain the solution $x=60/11$.
My question is however can one prove without analysis that x=60/11 of $BD$ in order to minimize the distance $\|ACE\|$?
 A: According to your math, you're minimizing $|AC|+|CE|$, not $|AC|+|CD|$.
Reflect $E$ about the line $BC$ to $E'$. Observe that $|CE| = |CE'|$, so the problem is equivalent to minimizing $|AC| + |CE'|$. Where should $C$ be? (Hint: The answer is not actually the midpoint.)
You can think of $ACE$ as a light ray reflected in the mirror $BD$. From one point of view, the solution to this problem is actually the reason why mirrors reflect light that way in the first place.
A: Rahul Narain's and robjohn's answers explained details well how to solve.
I just wanted to add a picture to show the steps of the answer. Reflect E across BD to E′. The shortest $|ACE'|$ should be line. It was shown as blue line in the picture below. Others are longer as you can see.
how to find the $|BC|=x$
$\triangle ABC$ is similiar to $\triangle E'DC $ 
Thus 
$$\frac{10}{12}=\frac{x}{12-x}$$
$$x=\frac{60}{11}$$

A: You must mean $|ACE|$.
Reflect $E$ across $BD$ to $E'$, then $|CE|=|CE'|$ and because a straight line is the shortest distance between two points, The shortest distance from $A$ to $C$ to $E'$ is for $C$ to be $\frac{5}{11}$ of the way from $B$ to $D$ which would make $|BC|=\frac{60}{11}$.
