Points $S$ and $T$ are on side $\overline{CD}$ of rectangle $ABCD$ such $\overline{AS}$ and $\overline{AT}$ trisects $\angle DAB$. If $CT = 3$ and $DS = 6$, then what is the area of $ABCD$?
I have drawn the picture but what can you do next? Are there similar triangles or ratios to do? This question is pretty urgent. It is really hard for me. Thanks for helping!
Hint: $m\angle DAB=90^\circ$, so $m\angle DAS=30^\circ=m\angle SAT=m\angle BAT$. See that $\tan30^\circ=\dfrac{1}{\sqrt{3}}=\dfrac{6}{DA}$, assuming that $S$ is the point on $CD$ closest to vertex $D$. Thus, $DA={2\sqrt{3}}$.