geometryquadrilateral
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?
The lengths don't determine the angles: for example with all sides equal you have a rhombus, where the angle at any vertex could be anything in the interval $(0, \pi)$.
As stated the problem can have many solutions.
For example:
Let the quadrilateral $ABCD$ such that $AD=DB=BC=x$ and $AC=\sqrt{3}x$. See figure 1.
Figure 1
The angle $b$ depends on $a$ as the expression:
$$b = \frac{a}{2}- \frac {\pi}{2}+ \arccos(- \frac{1}{2 \sin {\frac{a}{2}}}+ \sin {\frac{a}{2}})$$
So we have many solutions, for example: $a=\frac{\pi}{3}$ and $b=\frac{\pi}{3}$ or $a=\frac{\pi}{4}$ and $b=\frac{\pi}{2}$ as shown in figure 2.
Figure 2
Best Answer
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 $x$.