One of the angles of a trapezoid $ABCD$ is $30^\circ$ and its diagonals are perpendicular to each other One of the angles of a trapezoid $ABCD (AB\parallel CD)$ is $30^\circ$ and its diagonals are perpendicular to each other. If the midsegment is $10$ and one of the bases is $8$, find the other base, the diagonals and the legs of $ABCD$.
Let $\measuredangle ABC=30^\circ$ and $MN$ be the midsegment $(M;N \in AD;BC)$. Let's $AB=a;DC=b;AD=c;BC=d.$ So $$MN=\dfrac{AB+CD}{2}\iff 10=\dfrac{a+8}{2}\iff20=a+8\iff a=12.$$ Clearly $a>b$.
I am not completely sure how we are supposed to interpret the fact that the diagonals of the trapezoid are perpendicular. I have tried the following: let $AC\cap BD=O;AO=x;OC=y;BO=z;OD=t.$ Then $$x^2+z^2=a^2=144\\y^2+z^2=d^2\\y^2+t^2=b^2=64\\x^2+t^2=c^2$$ From here we can derive $$a^2+b^2=x^2+y^2+z^2+t^2=c^2+d^2\\c^2+d^2=144+64=208$$
Now we are searching for another relationship between $c$ and $d$. I wasn't able to find such.
Yet let $CC_1\perp AB,DD_1\perp AB$. Since $\measuredangle C_1BC=30^\circ\Rightarrow CC_1=DD_1=h=\dfrac{d}{2}$. Now I decided to see what happens if $AD_1=x\Rightarrow C_1B=4-x$. Nothing helpful, though.

Hint

The triangles $\Delta OCD$ and $\Delta OAB$ are similars with raio $8/12 = 2/3$. It means we can write $OD=2x$, $OB=3x$, $OC=2y$ and $OA=3y$.
We can also write $$\tan (OBA) = \frac yx \quad (1)$$ and $$\tan (OBC)=\tan (30-OBA)=\frac{\tan (30)-\tan(OBA)}{1+\tan(30)\tan(OBA)}=\frac 23 \frac yx.\quad (2)$$
Solve the system with equations $(1)$ and $(2)$, and find $y/x$.
You also know that $$4x^2+4y^2=64.$$
Can you finish?
EDIT:
Setting $\tan(OBA)=y/x=k$ on equation $(2)$
$$\frac{\sqrt(3)/3-k}{1+\sqrt(3)/3 k}=\frac 23 k.$$
