if such three condition find the $\sin{\angle OPA}$ Convex quadrilateral $ABCD$,and circumcenter is $O$,if  Point $P$ lie on sides $AD$,and such
$$\dfrac{AP}{PD}=\dfrac{8}{5},~~PA+PB=3AB,~~PB+PC=2BC,~~PC+PD=\dfrac{3}{2}CD$$find $\sin{\angle OPA}$

I try let $AB=a,BC=b,CD=c,DA=d,,AP=x,PD=y,x+y=d$,and $OA=1$,then I get very ugly
use this wiki
$$1=R=\dfrac{1}{4}\sqrt{\dfrac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$$
where $s$ is semiperimeter.
I want use this identity
$$\cos{(\angle APB+\angle CPB+\angle CPD)}=0$$
or $\angle APB=\angle 1,\angle CPB=\angle 2,\angle CPD=\angle 3$
$$\cos{\angle 1}\cos{\angle 2}\cos{\angle 3}=\sin{\angle 1}\sin{\angle 2}\cos{\angle 3}+\sin{\angle 1}\sin{\angle 3}\cos{\angle 2}+\sin{\angle 2}\sin{\angle 3}\cos{\angle 1}$$
 A: I have used coordinate geometry to attach this problem.  
Use an unit circle at the original $x^2 + y^2 = 1$ and the line y = -a.  In this way we have :
$A(-\sqrt{1 - a^2}, -a)$ and $D(\sqrt{1 - a^2}, -a)$
Using the ratio formula $P(\frac{3\sqrt{1 - a^2}}{13}, -a)$
Using distance formula we have
$PA = \frac{3}{13}\sqrt{1 - a^2} + \sqrt{1 - a^2} = \frac{16}{13}\sqrt{1 - a^2}$
$PD = \sqrt{1 - a^2} - \frac{3}{13}\sqrt{1 - a^2} = \frac{10}{13}\sqrt{1 - a^2}$
Also let B be $(x_1, \sqrt{1 - x_1^2})$ and C be $(x_2, \sqrt{1 - x_2^2})$
Hence $AB = \sqrt{(x_1 + \sqrt{1 - a^2})^2 + (\sqrt{1 - x_1^2} + a)^2}$
$PB = \sqrt{(x_1 - \frac{3}{13}\sqrt{1 - a^2})^2 + (\sqrt{1 - x_1^2} + a)^2}$
$BC = \sqrt{(x_1 - x_2)^2 - (\sqrt{1 - x_1^2} - \sqrt{1 - x_2^2})^2}$
$PC = \sqrt{(x_2 - \frac{3}{13}\sqrt{1 - a^2})^2 + (\sqrt{1 - x_2^2} + a)^2}$
$CD = \sqrt{(x_2 - \sqrt{1 - a^2})^2 + (\sqrt{1 - x_2^2} + a)^2}$
Put the above distances into PA + PB = 3AB theoretically we can express $x_1$ in terms of a.
Similarly for PC + PD = $\frac{3}{2}CD$ We also can express $x_2$ in terms of a.
By putting $x_1$ and $x_2$ into PB + PC = 2BC the equation can be solved for a.
Once $a$ is found we can find the answer :
$$\sin\angle OPA = \frac{13}{\sqrt{\frac{9}{a^2} + 160}}$$
The 3 equations are too difficult to solve and can only be resolved by tries and error.
My estimation gives the following results :
$a = 0.1$, $x_1 = -0.75$, $x_2 = 0.35$, $\sin \angle OPA = 0.399$.
