A trapezoid has diagonal from top left to lower right. The median is crossed by the diagonal. The length of the left part of the median is $x$, the length of the right part is $4.2$. The length of the bottom (longer) base is 12 and the length of the top base has been shown to be $2x-3.6$. Edit: The diagonal is bisected by the median.
How do I find $x$?
Let the median be $PQ, P \in AD, Q \in BC$ and $PQ \bigcap BD=T$. Using Thales's Theorem for $\triangle ABD$ and the line $PT$ gives:
$$\frac{x}{12}=\frac{DT}{BD}$$
Thales's theorem for $\triangle BCD$ and $QT$:
$$\frac{4.2}{2x-3.6}=\frac{BT}{BD}$$
Now note that the $2$ fractions on the right sum up to $1$ and you can solve for $x$