let $AC$and$AB$ intersect the tangent to the semicircle at $N$ at $K$ and $L$ respectively, let $AN$intersect $CD$ at $T$ where $D$ is the point on $AB$ such that $CD$ bisects $\angle ABC$ internally. by similar triangles, $NL=KL-KN=\frac{AK}{AC}\cdot BC-\frac{BC}{2}=\frac{BC(BC+AC)}{2AC}$ and by law of sine we have $\frac{sin(\angle BAN)} {NL}=\frac{sinB}{AN}=sinB\cdot\frac{sin\frac{\pi}{2}}{AN}=\frac{sin(\angle NAC)}{NK}$ so $\frac{sin(\angle BAN)} {sin(\angle NAC)}=\frac{BC+AC}{AB}$. and also we have $\frac{sin(\angle BAN)}{DT}=\frac{sin(\angle DTA)}{AD}=AC\cdot AD\cdot \frac{sin(\angle CTA)}{AC}=AC\cdot AD\cdot \frac{sin(\angle NAC)}{CT}$ by the definition of $D$, it follows that $AD=\frac{AB\cdot AC}{BC+AC}$ and we're done.