There is an Equilateral trapezoid $ ABCD. $ $ \bar{AB} =60, \bar{BC}=\bar{DA}= 36, \bar{CD}=108. $ $ M $ is the middle point of $ \bar {AB} $, and point $P$ on $ \bar{AM} $ follows that $ \bar {AP} $ =10. The foot of perpendicular dropped from $P$ to $ \bar {BD} $ is $E$. $ \bar{AC} \cap \bar{BD} $ is $ F $. Point $X$ is on $ \bar {AF} $ which follows $ \bar{MX}=\bar{ME} $ Find $ \bar{AX} \times \bar{AF} $