An equilateral triangle is drawn in a circle with one of its vertices on the diameter. What is $x$ in the figure?

In triangle $CAF$ $\angle AFC=\angle FCA=45^\circ$, hence $AO\perp FC$ and

\begin{align} \triangle ABO:\quad a&=\frac 6{\cos 75^\circ} =6(\sqrt2+\sqrt6) ,\\ x&=a\sin15^\circ\cdot \sqrt2 = 6(\sqrt2+\sqrt6)\cdot\tfrac{\sqrt2}4\,(\sqrt3-1)\cdot \sqrt2 =6\sqrt2 \approx 8.485281372 . \end{align}

Denote $AB=AD=BD=a$ and $OA=OC=r$. Note that $AC=\sqrt 2r$, so $a=\sqrt 2 r-x$. Consider $\triangle ABC$. By law of sines we have: $$\frac{AB}{\sin 45^\circ}= \frac{AC}{\sin 75^\circ}= \frac{BC}{\sin 60^\circ},$$ i.e. $$\frac{\sqrt 2r-x}{\sin 45^\circ}= \frac{\sqrt 2r}{\sin 75^\circ}= \frac{r+6}{\sin 60^\circ}.$$ Recall $\sin 45^\circ=\frac{\sqrt 2}{2}$, $\sin 60^\circ=\frac{\sqrt 3}{2}$ and $\sin 75^\circ=\frac{\sqrt 2+\sqrt 6}{4}$. Now first calculate $r$ from the equality of the second and the third term, and then calculate $x$.