Finding the fourth side given other three and diagonals.

Question

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?

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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?

Answer 1

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 $x$.

Answer 2

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 $x$.

Answer 3

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.