In $\triangle ABC$ we have $AB=7, AC=8, BC=9$. Point $D$ is the midpoint of the arc $BC$ of the circumcircle of $\triangle ABC$. Compute $\displaystyle\frac{AD}{BD}$, $BD$, and $CD$.
This is what I have so far but I am unsure if what I have is correct:
Since arc$BD$ = arc$CD$, $BD=CD=x$.
Using Ptolemy's Theorem in $ABDC$ we obtain $AB \cdot CD + AC \cdot BD = BC \cdot AD$. So $7 \cdot x + 8 \cdot x = 9 \cdot AD$ and $15x=9AD$.
Since $x=BD=CD$, $15BD = 9AD$ and $\displaystyle\frac{AD}{BD} = \frac{15}{9} = \frac{5}{3}$.
So this would mean that $BD=CD=3$. 
$AD$ is the angle bisector of $\widehat{BAC}$, hence if we call $X$ the intersection of $AD$ and $BC$, we have $BX=\frac{7}{15}\cdot 9$ and $CX=\frac{8}{15}\cdot 9$ by the bisector theorem. By Stewart's theorem the length of $AX$ is given by $$ AX^2 = \frac{7\cdot 8}{(7+8)^2}((7+8)^2-9^2) $$ hence $AX=\frac{8}{5}\sqrt{14}$ and since $AX\cdot XD=BX\cdot XC$ we have $XD=\frac{9}{10}\sqrt{14}$, from which $AD=\frac{5}{2}\sqrt{14}$. By Ptolemy's theorem it follows that $15\cdot BD = AD\cdot BC = \frac{45}{2}\sqrt{14}$, hence $$ BD=CD=\frac{3}{2}\sqrt{14},\qquad \frac{AD}{BD}=\frac{5}{3}. $$