Isosceles triangle has the least perimeter among triangles on the same base with same area? 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.
 A: Consider the diagram:

Here the line segments AD, EF and GC are parallel.
The triangles ABD and AHD have the same area as they have the same base and height.
The triangle AHD is isosceles as AH = DH through symmetry or Pythagoras.  Note that H is the midpoint of EF.
Now the distance AB+BC is the same as the distance AB+BD by reflection symmetry.
This distance must be smaller than the distance of the straight line AH+HC=AH+HD.
To conclude the isosceles triangle must have the smallest perimeter as the point B is arbitrary.
Hope this helps.
A: Let assume that base has endpoint as $(0,0)$ and $(x,0)$ and height $y$ ie $(x_0,y)$ where $0\le x_0 \le x$.
Now for different $x_0$, the perimeter will change without changing the area, therefore lets calculate the general perimeter.
the base is common, so lets just ignore it. hence the length of other two side is 
$x_0^2 + y^2 + (x - x_0)^2 +  y^2 = 2y^2 + x^2 + (x - x_0)^2 = f(x_0)$
Now lets maximize the $x_0$ by differentiating it w.r.t to $x_0$. Therefore 
$f'(x_0) = 4x_0 -2x$ or $4x_0 = 2x$
implies $x_0=\frac x2$.
Now you could do second derivative to check if its indeed mini ma or arbitrary substitute any value of $x_0$ other than $\frac{x}{2}$ and see if $f(x)$ is bigger than if it were $\frac x2$. If you check the figure, it satisfies the definition of being an isosceles triangle.
A: 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.
