Solution:

Points $D, E$ lie beyond $B$ on ray $AB,$ and $F$ lies on the opposite side of line $AB$ from $C.$ Since $ACFD$ and $BCFE$ are cyclic, the inscribed angles give $\angle FDA = \angle FCA$ and $\angle FEB = \angle FCB.$ Writing $\alpha = \angle FCA$ and $\beta = \angle FCB,$ triangle $DEF$ has angles $\angle FDE = 180^\circ - \alpha$ and $\angle FED = \beta,$ so $\angle DFE = \alpha - \beta = \angle ACB.$ From triangle $ABC,$ $\cos \angle ACB = \frac{25 + 36 - 16}{2 \cdot 5 \cdot 6} = \frac{3}{4},$ so the law of cosines in triangle $DFE$ gives $$DE^2 = 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cdot \tfrac{3}{4} = 32, \qquad DE = 4\sqrt{2}.$$

In triangle $DFE,$ $\cos \angle FDE = \frac{4 + 32 - 49}{2 \cdot 2 \cdot 4\sqrt{2}} = -\frac{13\sqrt{2}}{32},$ so $\alpha$ is acute with $\cos\alpha = \frac{13\sqrt{2}}{32}$ and $\sin\alpha = \sqrt{1 - \frac{338}{1024}} = \frac{7\sqrt{14}}{32}.$ Let $G$ be the intersection of line $CF$ with line $AB.$ In triangle $ACG,$ $\angle GAC = \angle BAC$ has $\cos \angle BAC = \frac{16 + 36 - 25}{2 \cdot 4 \cdot 6} = \frac{9}{16},$ $\sin \angle BAC = \frac{5\sqrt{7}}{16},$ and $\angle ACG = \alpha,$ so $\sin(\angle BAC + \alpha) = \frac{5\sqrt{7}}{16} \cdot \frac{13\sqrt{2}}{32} + \frac{9}{16} \cdot \frac{7\sqrt{14}}{32} = \frac{\sqrt{14}}{4}$ and $$AG = \frac{AC \sin \alpha}{\sin(\angle BAC + \alpha)} = \frac{6 \cdot \frac{7\sqrt{14}}{32}} {\frac{\sqrt{14}}{4}} = \frac{21}{4}.$$

Line $CF$ is the radical axis of the two circles, so $GA \cdot GD = GB \cdot GE.$ With $x = BD$ and $BE = x + DE = x + 4\sqrt{2}:$ $$\frac{21}{4}\left(x - \frac{5}{4}\right) = \frac{5}{4}\left(x - \frac{5}{4} + 4\sqrt{2}\right),$$ since $GD = 4 + x - \frac{21}{4}$ and $GB = \frac{21}{4} - 4.$ This gives $16\left(x - \frac{5}{4}\right) = 20\sqrt{2},$ so $x = \frac{5 + 5\sqrt{2}}{4}$ and $BE = \frac{5 + 21\sqrt{2}}{4}.$ Therefore $a + b + c + d = 5 + 21 + 2 + 4 = 32.$