Three different points $A,B$ and $C$ lie on a circle with center $M$ so that $| AB | = | BC |$. Point $D$ is inside the circle in such a way that $\vartriangle BCD$ is equilateral. Let $F$ be the second intersection of $AD$ with the circle . Prove that $| F D | = | FM |$.
$\angle FDC=\pi-\angle BCD-\angle ADB=\frac{2\pi}{3}-\angle FAD=\frac{2\pi}{3}-(\pi-\angle FCB)=\angle FCD$ so $\Delta FCD$ is isosceles. Furthermore $\angle CFD=\pi-\angle B=2\angle A=\angle BMC$, and together with $DC=BC$ we get $DCF$ is congruent to $BMC$.
Thus $FM=BM=FD$