Denote $b=\frac{\angle BAC}{2},\ c=\frac{\angle BCD}{2}$ and assume WLOG $\angle ABC\le\angle BCD$. Let $F$ be the intersection of the bisector of $\angle DAB$ and $\angle CDA$. Take $E$ on segment $BC$ such that $BE=AB$. By the given condition it clearly exists and $CE=CD$. Because $ABCD$ is a cyclic quadrilateral $$\angle BAE=90^\circ-b,\ \angle BAF=\frac{180^\circ-\angle BCD}{2}=90^\circ-c\implies \angle FAE=c-b$$.Since $c>b$ point $F$ lies inside angle $BAE$. Analogously we obtain $$ \angle FDE=c-b.$$Points $E,F$ lie on the same side of line $AD$ so quadrilateral $DAFE$ is cyclic. Hence$$\angle DEF=180^\circ-\angle FAD=90^\circ+c=180^\circ-\angle DEC\implies F\in CE^\rightarrow\implies F\in\overline{BC}.$$