Position of Q in which the area of triangle CDQ is a third part of the area of square ABCD. Given


*

*$A(8,0)$

*$B(0,4)$

*$C(-4,-4)$

*$D(4,-8)$

*$P(2,3)$


Question
In the figure the triangle $CDQ$ is grayed out. There is a position of $Q$ in which the area of triangle $CDQ$ is a third part of the area of square $ABCD$. For this position, calculate exactly the coordinates of $Q$.
Answer:
I calculates that the height of triangle $CDQ$, with base $CD$, must be $2/3$ part of the side of the square. 
I became step one of the solution, but I do not understand the following steps of the answer. Can someone help? I added a picture of the answer! 



The side of the square is $4\sqrt{5}$, so the area of the square is $80$. 
The area of $\triangle_{QCD}=\frac{1}{2}\cdot4\sqrt{5}\cdot h=\frac{80}{3}$, so the height $h=\frac{8\sqrt{5}}{3}$, which is $\frac{2}{3}$ the side of the square. Now imagine drawing a line through the midpoint of $CD$, say $R$, through $Q$, to the midpoint of $AB$, say $S$, this bisects the square and has $RQ:QS=\frac{2}{3}:\frac{1}{3}$, or equivalently $RQ:QS=2:1$, and so $DQ:QP=2:1$ also.
Now find $\overrightarrow{DP}=\overrightarrow{DO}+\overrightarrow{OP}=-\overrightarrow{OD}+\overrightarrow{OP}=\binom{-4}{8}+\binom{2}{3}=\binom{-2}{11}$.
You can now find $\overrightarrow{OQ}=\overrightarrow{OD}+\frac{2}{3}\overrightarrow{DP}=\binom{4}{-8}+\frac{2}{3}\cdot\binom{-2}{11}=\binom{8/3}{-2/3}$, giving $Q=(\frac{8}{3},-\frac{2}{3})$ as required.
