F is the midpoint of BC in parallelogram ABCD. AF intersects BD in point E, CE intersects AB in point K. KB = 5, AD = 12. Angle A equals 30 degrees How can I find the area of ABCD according to that?
The main problem is: how can I find AB? If I find it, I will know the height.
Construct $AC$. Let the point where $BD$ intersects $AC$ be $G$.
By Ceva's Theorem:
$$ \frac{AK}{KB} \frac{BF}{FC} \frac{CG}{GA} = 1 $$
Since $BF = FC$ and $CG = GA$ (diagonals of parallelograms bisect each other), $AK = BK$
I assume you can continue from here.
Joining $AC$, crossing $BD$ at $G$ (see @Benson Lin), then since $BG$ and $AF$ are median lines in triangle $ABC$ crossing at $E$, and $CK$ is drawn through $E$, therefore $CK$ is also a median line and $AK = KB = 5$. And since the angle of the parallelogram at $A = 30$ degrees, the perpendicular from $B$ to $AD = \frac12AB = 5$. Thus the area of $ABCD = 12*5 = 60$.