Quadrilateral

Join $BD$ and $AC$ . $\angle ABD=90^{\circ},\angle ACD=90^{\circ}$ (Angle subtended at the semicircle)

Now by pythagorean theorem , we get $BD=\sqrt{4^2-1^2}=\sqrt{15}$ and $AC=\sqrt{4^2-CD^2}=\sqrt{16-CD^2}$ .

Now by ptolemy's theorem , $AB \cdot CD+BC \cdot AD=AC \cdot BD$ . or $1 \cdot CD+ 1 \cdot 4 = \sqrt{16-CD^2} \cdot \sqrt{15}=\sqrt{240-15CD^2}$ .

Let $CD=x$ . We have $x+4=\sqrt{240-15x^2}$ or $x^2+8x+16=240-15x^2$ or $16x^2+8x-224=0$ or $2x^2+x-28=0$ .

By quadratic formula we have $x=\dfrac{-1 \pm \sqrt{1^2-4 \times 2 \times (-28)}}{4}=\dfrac{-1 \pm 15}{4}=\dfrac{14}{4}=3.5$ (we have to consider only positive value as length of side cannot be negative).

Hence $x=CD=\boxed{3.5}$ .

Alternative Proof:

Let $\angle AOB=\theta.$ Therefore $\angle COD=\pi-2\theta\Rightarrow CD=4sin$ ( $\frac{\pi-2\theta}{2})$ $=4cos\theta=4.\frac{(OA^2+OB^2-AB^2)}{2.OA.OB}=\frac{7}{2}=\boxed {3.5}.$