Cyclic Quad Lengths

Relevant wiki: Similar Triangles - Problem Solving - Medium

Figure shows

$\triangle EBC \sim \triangle FDB$ , then
$EB : FD =EC : FB \Leftrightarrow EB\times FB = EC \times FD \qquad \qquad (1)$

$\triangle ACF \sim \triangle ABE$ , then
$AC : AB =FC : EB \Leftrightarrow EB\times AC= FC \times AB \qquad \qquad (2)$

From $(1)$ and $(2)$ ,

$\dfrac{EB^2 \times FB \times AC}{EC \times FC \times FD} = AB = \boxed{10} \; .$

By Pascal's theorem on $AABBDC$ we have that $EF$ is parallel to the tangent at $A$ .

Let $X$ be an arbitary point on the tangent from $A$ on the opposite side to $D$ . We have $\angle FEC=180-\angle XAC$ but also $\angle FBC=180-\angle CBA=\angle CDA=\angle XAC$ . This means that $\angle FEC+\angle FBC=180$ so $BCEF$ is cyclic.

By repeated application of intersecting chords and tangent secant on the two circles we get:

$\frac{EB^2 \times FB \times AC}{EC \times FC \times FD}=\frac{EC \times AE \times FB \times AC}{EC \times FC \times FD}=\frac{AE \times AC \times FB}{FC \times FD}=\frac{AB \times AF \times FB}{FC \times FD}=\frac{AB \times FC \times FD}{FC \times FD}=AB$

So therefore the given expression is $8\times AB=8 \times 10=80$ .