I'd like some help with the following Olympiad Problem about a trapezoid:
There is a trapezoid $ABCD$ with parallel sides $BC$ and $AD$ such that $AB=1$, $BC=1$, $CD=1$ and $DA=2$. Let $M$ be the midpoint of $BC$. Suppose there are two points $E,F$ on $AD$ and a point $G$ on $CD$ such that $\angle EMG = \angle FGM = 90^{\circ}$ and $EF = \frac{3}{2}$. Find the length of the segment $DG$.
As of currently, I've drawn the following diagram:
The issue from here is that I have no clue where to go xD I know the answer is $\frac{7-\sqrt{33}}{8}$, but the method to get there, I need help with.
Any help would be greatly appreciated!
You might go like this: $$A=(-1;0)$$ $$B=(-\frac12;\frac12\sqrt3)$$ $$M=(0;\frac12\sqrt3)$$ $$C=(\frac12;\frac12\sqrt3)$$ $$D=(1;0)$$ Then consider a general point $G(g;f(g))$ on $CD$ (the latter allows to give a term for $f(g)$). From $MG$ you now can calculate both the searched for distance (still depending on $g$) as well as the linear equation of that very line.
Next set up the line equations for $ME$ and $GF$ by using the points $M$ and $G$ respectively, while the inclination would be the negative inverse to that of the former. The zero-height points of these lines then are just $E$ and $F$ respectively. I.e. those can be calculated, still depending on the parameter $g$. However, their distance $d(E,F)=\frac32$ is given, thus this now allows to solve for $g$, which finally can be entered into the above already pre-calculated, so far parametrized distance value.
--- rk