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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.

enter image description here

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    $\begingroup$ The classical proof is to reflect point B about the line $C'C$. Then It's easy to show that $AB'$ is a straight line and it goes thru $C'$. $\endgroup$ – Vasya Mar 30 '20 at 12:53
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Perhaps look at this picture ...

enter image description here

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  • $\begingroup$ So the bold red line represents $AC+BC$, which is minimal as a straight line? $\endgroup$ – Nεo Pλατo Mar 30 '20 at 13:03

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