For reference: A line tangent at $B$ to the circle circumscribed to the triangle $ABC$, is parallel the interior bisector $CD$($D$ in $AB$). Calculate $AC$, if $AD = 5$ and $BD=4$ (Answer:$7,5$)
My progresss: I believe the figure is this
Th. Angle Bissector:$\triangle ABC$
$\frac{AC}{5} =\frac{BC}{4}\implies AC = \frac{5BC}{4}\\ CD^2 = AC.BC - 5.4 \implies \\ AC =\frac{CD^2 +20}{BC}\\ \frac{5BC^2}{4}=CD^2 + 20 \implies CD = \frac{\sqrt{5BC^2-80}}{2}$
missing a relationship with BC...???
$BC$ subtends $2 \angle A$ at the center of the circlec, say $O$.
Then $\angle OBC = 90^\circ - \angle A$ and hence the tangent makes angle $\angle A$ with segment $BC$ but given the tangent is also parallel to the angle bisector, we conclude $\angle A = \angle C / 2$.
That leads to $\triangle ADC$ being isosceles and hence $~ CD = 5$.
Now plug this into your work to find $AC$.