Circles and Quadrilaterals

A convex quadrilateral $ABCD$ exists such that the angle bisector at $A$ passes through $C$ . The extensions of $BC$ and $AD$ meet at $E$ such that $\\$ $\angle CED = \angle CDA - \angle BAD$ . $\\$ Let $\omega$ be the tangent to the circumcircle of $\triangle BCD$ at $D$ , and let the angle that $\omega$ makes with $CD$ be $\alpha$ . $\\$ Then $\angle BAD$ = k $\alpha$ . $\\$ Find k.