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 $\bar{AX} \times \bar{AF}$