Finding the fourth side given other three and diagonals. Say I had a quadrilateral and was given sides of length $a,b,c$ and diagonals $d,e.$ How can I find the fourth side?

Let the fourth side be $x$.
I though of using Ptolemy's first of all to get
$$ac + cx \ge de.$$
We can solve for $x$ to get
$$\frac{de-ac}{c} \le x.$$
But, that doesn't really help much, other than giving a bound. I can't really think of another good way to do this. I also thought of using the Law of Cosines somehow. Can someone help?
 A: Label the vertices $P, Q, R, S$ so that $PQ = a, QR = b, RS = c, SP = x, PR = d$, and $QS = e$.
Since you know all three sides of $\Delta PQR$ you can find the angles using the Law of Cosines.  Likewise for $\Delta QRS$. So you can find $\angle PQR$ and $\angle SQR$. Subtracting gives you $\angle PQS$. Using the Law of Cosines one more time gives you $x$.
A: Here is one way to approach it.

Say we know $a, b, c, d, e$ and we need to find $x$.
Apply law of cosine in $\triangle BAD$ to find $\angle A$.
Then apply law of cosine in $\triangle CAD$ to find $\angle CAD$.
Now you know $\theta$. Apply law of cosine in $\triangle BAC$ to find $x$.
A: Hint. You can use coordinate geometry to find the endpoints of the third side, each of which is determined by the triangles whose sides are $a, b, d$ and $a, c, e$ respectively. The $x$ is the distance between the two points. Clearly you should be able to do this once you've chosen an origin, a unit and an orientation of axes, since the sides with lengths $b, d$ and $c, e$ respectively in each triangle determine intersecting lines, which give you the needed points.
