An isosceles trapezoid $ABCD$ function of $AB$ Ok so i have an isosceles trapezoid $ABCD(AD=BC)$ with bigger base $AB$ and let $O$ be the point of the crossed diagonals and let $OH$ be perpendicular to $AB$.
I want to find a function for $AB$ with variables $AD$, $BC$, $DC$, and $OH$.
$f(AD,BC,DC,OH) = AB$
$f = ???$
 A: Define $p = \frac{1}{2}|\overline{AB}|$, $q = \frac{1}{2}|\overline{CD}|$, $s = |\overline{AD}| = |\overline{BC}|$, $h = |\overline{OH}|$. Let $K$ be the foot of the perpendicular dropped from $C$ to $\overline{AB}$, and define $k = |\overline{CK}|$.

Observe that, from right triangle $\triangle BKC$, we have $k^2 + (p-q)^2 = s^2$.
Now, since $\triangle AOH$ and $\triangle ACK$ are similar:
$$\frac{|\overline{OH}|}{|\overline{AH}|} = \frac{|\overline{CK}|}{|\overline{AK}|} \qquad \to \qquad \frac{h}{p} = \frac{k}{p+q} \qquad \to \qquad \frac{h^2}{p^2} = \frac{s^2 - (p-q)^2}{(p+q)^2}$$
Therefore,
$$p^4 - 2 p^3 q + p^2 \left( q^2 - s^2 + h^2 \right) + 2 p q h^2 + h^2 q^2 = 0$$
The fourth-degree polynomial in $p$ doesn't factor, so the roots are tricky to find. There's a Quartic Formula, but it's exceedingly messy; given specific values for $q$, $s$, and $h$, numerical approximation methods may be your best bet. In any event, once you have the desired root $p$, you can write
$$|\overline{AB}| = 2 p$$
Note: By the Descartes Rule of Signs, the polynomial has either two positive roots for $p$, or none. So, you may have a choice to make, which could preclude there being a single-valued function for the value of $|\overline{AB}|$. 
