The original problem is this:
The base $ABCD$ of a pyramid $ABCDM$ is inscribed in a circle, $AB = 9$, $BC = 10$ and $\angle ABC = 60^\circ$.
Find the altitude of the pyramid if all dihedral angles at the base are equal to $60^\circ$.
I understand that $\angle ADC = 120^\circ$ (sum of opposite angles equal to $180^\circ$) and that because the dihedral angles are equal a circle can be inscribed as well, making it a Bicentric Quadrilateral, but this is as far as my knowledge goes.
I could also find $AC$ if I want, but I don't see how that helps us.
If $AD=x$ then $CD=x+1$ and: $$ AC^2=9^2+10^2-2\cdot10\cdot9\cos60°=x^2+(x+1)^2-2x(x+1)\cos120°. $$