What's the measure of the segment $PD$ in the triangle below?

For reference:- In the triangle $ABC$: $AB = 7, BC = 9$ and $\angle B = 74^o$. The angle bisector of B intersects the perpendicular bisector of AC in P. Calculate $PD$; if $AC \cap BP = D$. (Answer:$3,7$)

My progress:

$P \in $ circumscribed circle $\triangle ABC \implies$

$ABCD$ is cyclic

$B$ is angle bissector $\implies AP = CP$

Ptolemy's Th:

$AB.CP+BC.AP = BP.AC \implies\\ 7CP+9AP = AC.BP \therefore \boxed{16CP = AC.BP}$

Angle Bissector Th.:

$\frac{AD}{7} = \frac{CD}{9} \implies \frac{AD}{CD} = \frac{7}{9}$

Using formula for the length of angle bisector

$BD^2 =7.9-AD.DC =63-AD.DC$

Solution 1:

Using power of point $D$, $AD \cdot DC = BD \cdot DP$. So your last equation can be rearranged to give $$BD \cdot BP = 63$$ from which it is easy to work out $DP=BP-BD$.