# How does an isosceles triangle minimize distance?

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.

• 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 $C'$. – Vasya Mar 30 '20 at 12:53

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