BC wanted

Draw the picture so that the chord $GH$ is horizontal. Then $E$ is vertically above the centre $O$ of the circumcircle of $ABCD$ . With the natural coordinate system, suppose that $E$ and $F$ have coordinates $(0,y)$ and $(x,y)$ respectively, and suppose that the circumcircle has radius $R$ . The circumcentre of the triangle $ADE$ is the midpoint $J\; (x,0)$ of $AD$ , and so $JA = JE$ , which implies that $\sqrt{x^2 + y^2} = \sqrt{R^2 - x^2}$ , so that $2x^2 + y^2 = R^2$ . But $144 = GF \times FH \; =\; \big(\sqrt{R^2 - y^2} + x\big)\big(\sqrt{R^2-y^2} - x\big) \; = \; R^2 - x^2 - y^2 \; = \; x^2$ and hence $x=12$ . On the other hand, simple angle-chasing (there are a number of similar right-angled triangles here) shows that $IBE$ and $ICE$ are both isosceles triangles, and therefore that $BI = CI = EI$ . The Intersecting Chords Theorem now tells us that $\begin{aligned}EI^2 & =\; BI \times CI \; = \; HI \times GI \; = \; \big(\sqrt{R^2 - y^2} + EI\big)\big(\sqrt{R^2 - y^2} - EI\big) \\ & = \; R^2 - y^2 - EI^2 \end{aligned}$ so that $EI^2 = \tfrac12(R^2 -y^2) = x^2$ , and hence $EI = x = 12$ . Thus $BC = 2EI = \boxed{24}$ .

Although it is not necessarily the case, we can assume $ABCD$ is a square and still fulfill the requirements of the problem (being inscribed in a circle and having perpendicular diagonals) and we should still arrive at the correct solution but with easier calculations.

Then $GH$ is a perpendicular bisector to $AD$ , so that $AF = FD$ , and by the intersecting chords theorem , $AF \times FD = GF \times FH$ , or $AF^2 = 144$ , which solves to $AF = 12$ . Therefore, $BC = AD = 2 AF = \boxed{24}$ .