I have been experimenting a bit with triangles and was wondering what general formulas exist for.
Given a triangle $\triangle ABC$ and a point $M$ inside the triangle. Let the side lengths of the triangle and two of the three connecting distances from one corner to $M$ be given. Is it possible to calculate the third connecting distance $\overline{CM}$, denoted by $x$, without using trigonometry/inverse functions? Does the calculation become easier if the triangle is special in some way, e.g. if $\angle ACB=\angle CBA$ applies?
There is a Heron-type formula for the volume of a tetrahedron, which gives the volume of a tetrahedron as a function only of its edges lengths $U$, $V$, $W$, $u$, $v$, $w$: $$ V = \frac{\sqrt {\,( - p + q + r + s)\,(p - q + r + s)\,(p + q - r + s)\,(p + q + r - s)}}{192\,u\,v\,w} $$ where \begin{align} p & = \sqrt {xYZ}, & q & = \sqrt {yZX}, & r & = \sqrt {zXY}, & s & = \sqrt {xyz}, \end{align} \begin{align} X & = (w - U + v)\,(U + v + w), & x & = (U - v + w)\,(v - w + U), \\ Y & = (u - V + w)\,(V + w + u), & y & = (V - w + u)\,(w - u + V), \\ Z & = (v - W + u)\,(W + u + v), & z & = (W - u + v)\,(u - v + W). \end{align} Here you have a tetrahedron of volume zero, hence one of the factors above the fraction is $0$. Solving for one edge length should hopefully give two answers, and the smaller is that of interest.
I don't have the courage yet to solve this myself; however there will be no trigonometry involved, I promise.