Help please. Really appreciate it

Let ABCD be a convex quadrilateral,

and let P, Q, R, S, T, U, V, and W be the trisection points of the sides of ABCD, as shown. 

If the area of quadrilateral ABCD is 180, then find the area of hexagon AQRCUV.

Let AB = a
Let BC = b
Let CD = c
Let DA = d

Let $\angle{ABC}$ = B
Let $\angle{UDV}$ = D

$\begin{array}{|rcll|} \hline Area_{\text{ABC}} &=& \frac{ab\sin(B)}{2} \\ Area_{\text{CDA}} &=& \frac{cd\sin(D)}{2} \\ Area_{\text{ABCD}} &=& Area_{\text{ABC}} + Area_{\text{CDA}} \\ &=& \frac{ab\sin(B)}{2} + \frac{cd\sin(D)}{2} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline Area_{\text{QBR}} &=& \frac{\frac{a}{3}\frac{b}{3}\sin(B)}{2} \\ Area_{\text{UDV}} &=& \frac{\frac{c}{3}\frac{d}{3}\sin(D)}{2} \\ Area_{\text{hexagon}} &=& Area_{\text{ABCD}} - Area_{\text{QBR}} - Area_{\text{UDV}} \\ Area_{\text{hexagon}} &=& Area_{\text{ABCD}} -\frac{\frac{a}{3}\frac{b}{3}\sin(B)}{2} - \frac{\frac{c}{3}\frac{d}{3}\sin(D)}{2} \\ Area_{\text{hexagon}} &=& \frac{ab\sin(B)}{2} + \frac{cd\sin(D)}{2} -\frac{\frac{a}{3}\frac{b}{3}\sin(B)}{2} - \frac{\frac{c}{3}\frac{d}{3}\sin(D)}{2} \\ Area_{\text{hexagon}} &=& \frac{ab\sin(B)}{2}\left(1-\frac19 \right) + \frac{cd\sin(D)}{2}\left(1-\frac19 \right) \\ Area_{\text{hexagon}} &=& \frac{ab\sin(B)}{2}\cdot \frac89 + \frac{cd\sin(D)}{2}\cdot \frac89 \\ Area_{\text{hexagon}} &=& \frac89 \cdot \left( \frac{ab\sin(B)}{2} + \frac{cd\sin(D)}{2} \right) \\ Area_{\text{hexagon}} &=& \frac89 \cdot Area_{\text{ABCD}} \\ Area_{\text{hexagon}} &=& \frac89 \cdot 180 \\ \mathbf{ Area_{\text{hexagon}} } & \mathbf{=} & \mathbf{160} \\ \hline \end{array}$

The area of hexagon AQRCUV is 160

Draw AC

Now, because BQ  =(1/3) BA and BR =(1/3)BC, Q and R  divide BA and BC proportionally...

Thus.....BQ/BA  =BQ/BC...and by Euclid,  QR is parallel to AC

And <BQR = < BAC    and <QBR  = <ABC

So, by AA  congruency ΔQBR ~ ΔABC

And since AB=3QB.....then the area of ΔABC  will be 3^2  = 9 times that of ΔQBR

By similar reasoning, we can show that the area of ΔADC  = 9 times that of ΔVDU

And  area of ΔABC +  area of ΔADC  = 180....so....

9 * area of ΔQBR + 9 * area of ΔVDU  = 180

9 *  [ area of ΔQBR +  area of ΔVDU]  = 180     ....so...

[ area of ΔQBR +  area of ΔVDU]  = 180/9  = 20

So......the area of hexagon AQRCUV  =  180  -  20  =  160

I like the trig approach, heureka......!!!!!