How does an isosceles triangle minimize distance?

Question

I thought of this while solving another problem which I hacked using calculus but seems much easier than it is.

Let me make a general case of it, or at least an example.

Construct a segment $AB=6\text{cm}$ and the locus of all points $C$ such that $\triangle ABC$ has an area of $12\text{cm}^2$.

What you would do is draw a perpendicular $AC$ from any point on $AB$ that is $4\text{cm}$ long and then draw a line parallel to $AB$ at $C$. That line is the locus of points.

Now extend the question. What point $C$ on that line minimizes the sum of length of segments $AC$ and $BC$?

In the question I deduced it differently since it was a question in coordinate geometry and I found an isosceles triangle. So how is it that $AC+BC$ is minimum when $AC=BC$.

I feel like there's a Euclid style proof of this but I can't really get one at the moment.

Here's the picture of the situation.

Answer 1

Perhaps look at this picture ...