I've seen it before - 3

Draw CH is perpendicular to AD, O is the center of the semi-circle. Call K is the center of CD. $\angle{HCD}=\angle{KOD} \Rightarrow$ $\frac{CK}{OD}=\frac{HD}{CD} \Rightarrow DH=\frac{1}{2}$ $BC=AD-2\times DH=8-1=\boxed{7}$

Extend hhe lines AB,DC so that we get an isosceles triangle with top P. Then the triangle OCD is similar to ADP, and it follows that AP=16, hence BP=14. The triangle PBC is similar to ADP, so that 14/16=BC/8 => BC=7.

Let the intersection of the red lines (as shown) be T. Then AT = BT = 4 (radii of circle). Now note that $\frac{AT}{AB}$ = 2 (#1). Next, extend AB and CD until they meet at point P. This means that: $\ \angle BAD = a \implies\angle BCD = 180 - a \implies\angle BCP = a = \angle CBP$ Triangles ABT, BCP and ADP are similar so CP = BP = 2 ${x}$ from (#1). Finally: $\frac{DP}{DA} = \frac{CP}{CB} \implies\ \frac{x+1}{2} = 4 \therefore\ x= 7$

ง่ายสาดดดด

$\Delta ADC$ is right triangle. Applying Pythagoras' theorem, $AC^2=AD^2-CD^2=8^2-2^2=60$

Now applying Ptolemy's Theorem,

$AD\times BC+AB\times CD=AC\times BD$

$\Rightarrow 8\times CD+2\times 2=AC\times AC=60$

$\Rightarrow CD = \fbox{7}$

let h be the height of the trapezium.In triangle ABO,(2/4) * sqrt(4 * 4^2-2^2)=(1/2) * h * 4 or h=(sqrt 15)/2.Triangle BOC is an isosceles triangle,where the h will be the median of the triangle,applying pythagorean theorem we get BC/2=sqrt[4^2-{(sqrt 15)/2}^2] or BC/2=7/2 or BC=7