The midsegment of an isosceles trapezoid is $4$, the diagonal is $4\sqrt{2}$ and the leg is $2\sqrt{5}$. Find the area and the bases.
Let $DD_1$ and $CC_1$ be the altitudes and $MN$ be the midsegment $(M\in AD;N\in BC)$. Since the trapezoid is isosceles, $AD_1=BC_1=\dfrac12 (AB-CD)$. Futhermore, $MN=\dfrac12 (AB+CD)=4.$ I do not see how to use the diagonal and the leg to find the bases. Thank you in advance! Any help would be appreciated!
Following out the useful hints of @random:
Since $AC_1=\frac12AB+\frac12CD=MN$, and by Pythagoras $AC^2=AC_1^2+CC_1^2$, then$$CC_1^2=32-16$$and$$CC_1=4$$Next, in triangle $CC_1B$, since$$CB^2=CC_1^2+C_1B^2$$we have$$C_1B^2=(2\sqrt 5)^2-4^2)=20-16=4$$
and$$C_1B=2$$
Therefore$$AB=MN+C_1B=4+2=6$$and$$DC=D_1C_1=AB-2C_1B=6-4=2$$and from the area formula for trapezoids we find the area is$$\frac{CC_1(AB+CD)}{2}=\frac{4(6+2)}{2}=16$$