# Not that angle again

**Geometry**Level 4

In parallelogram \(ABCD\), \(F\) and \(E\) are on \(\overline{DC}\) and \(\overline{BC}\) respectively, such that \(\overline{AE}\) and \(\overline{AF}\) are heights. Suppose that \(CE=6\) and \(CF=3\) and that \(\angle{EAF}=60^{\circ}\). Then the area of \(ABCD\) is \(a\sqrt{b}\) where \(b\) is not divisible by the square of any prime. Find \(a+b\).