Prove that of all triangles on the same base with same area, the isosceles triangle has the least perimeter (without trigonometry).
I could prove this with trigonometry but couldn't do the same with elementary geometry. I can see that if AB is the common base, then the locus of the third point must be parallel to AB. I cannot understand anything more hence couldn't provide anything more. Just trying to solve a trigonometric problem with geometry. Thanks, in advance, for your solutions.
Best Answer
Draw line segment $AB$, representing the base. Since the area is fixed, this means that the height $h$ is fixed. Draw a line $L$ parallel to $AB$ that is $h$ units above $AB$. We seek a point $C$ on $L$ that minimizes the distance $AC + CB$.
Now imagine travelling from $A$ to $C$, but then instead of turning back around to $B$, we take the mirrored path (reflected in the line $L$) and arrive at a point $B'$, which is $h$ units above the line $L$ (and thus $2h$ units above $B$). Notice that $CB = CB'$. So it remains to find $C$ such that $AC + CB'$ is minimized.
But the shortest distance between any two points is a straight line! Notice by construction that $L$ bisects the straight line from $A$ to $B'$. We conclude that taking $C$ to be the midpoint (so that the triangle is isosceles) will minimize the perimeter.