Points $S$ and $T$ are on side $\overline{CD}$ of rectangle $ABCD$ such that $\overline{AS}$ and $\overline{AT}$ trisect $\angle DAB$. If $CT = 2\sqrt{3}-3$ and $DS = 1$, then what is the area of $ABCD$?
https://latex.artofproblemsolving.com/c/c/e/cce3c90dcfa09daf5610e4efbe8d333bcd0dafb5.png
Angle SAD must = 30°
So...triangle SAD is a 30-60-90 right triangle....and if DS = 1 ....then AD = sqrt (3) * DS = sqrt (3)
And AS= 2
And angle AST = 120° And angle TAS = 30° so angle ATS = 30°
So....AS = TS = 2
So CD = CT + TS + SD = [ 2sqrt (3) - 3 ] + [ 2 ] + [1] = 2sqrt (3)
So [ ABCD ] = AD * CD = 2sqrt (3) * sqrt (3) = 6 units^2