Trapezoid Root Mean Square I'm trying to prove that length of the line $AB$, parallel to both bases of a trapezoid, that cuts a trapezoid into two trapezoids of equal area is the Root Mean Square of the bases. In other words, if the length of the top base is $a$ and the length of the other base is $b$, then :
$AB=\sqrt{\frac{a^2 + b^2}{2}}$
I've been stuck on this for a couple of days. Can someone give me a hint? Please don't just state the proof. I've already tried using the areas of the two trapezoids to get the RMS. 
 A: Suppose that $a$ is the shorter (and "upper") base. Draw a line through an endpoint of the $a$-base, parallel to the opposite lateral side; this cuts the trapezoid into a parallelogram (with upper and lower bases $a$) and a triangle (with (lower) base $b-a$). The drawn line separates your "middle base", say of length $c$, into a part (of length $a$) within the parallelogram and a part (of length $c-a$) within the triangle.

If $h$ is the height of the parallelogram (and triangle), and $k$ is the height the upper sub-trapezoid (and upper sub-triangle), then, by similarity,
$$\frac{k}{h}=\frac{c-a}{b-a} \tag{1}$$
Now, assuming the middle base bisects the area, so that the upper sub-trapezoid's area is half the whole, we have
$$\frac{1}{2}k(a+c) = \frac12\cdot\frac12h(a+b) \quad\to\quad \frac{k}{h} = \frac12\frac{a+b}{a+c} \tag{2}$$
Setting $(1)$ and $(2)$ equal gives
$$\frac{c-a}{b-a}=\frac12\frac{a+b}{a+c}\quad\to\quad 2(c^2-a^2) = b^2-a^2 \quad\to\quad 2c^2 = a^2+b^2 \tag{3}$$
which is the target relation. $\square$
