Lines $DA$ and $CB$ meet at $E$, show that the radical center of the circles with diameters $AC$, $BD$, $CD$ is the orthocenter of $\triangle CED$ 
$ABCD$ is a quadrilateral such that the sides $DA$ and $CB$ produced intersect in $E$. Prove that the radical center of the circles by taking $AC$, $BD$ and $CD$ as diameters is the orthocenter of the triangle $CED$.

I attempted by dropping perpendicular from $C$ to $DA$ and from $D$ to $CB$ 
 (see the picture below), but I do not get anything further.

 A: 
Take a look at circle with diameter CD. It meets the line $AD$ at $F$ and line $AC$ at G. Obviously:
$$\angle CFD=\angle CFA=90^\circ$$
It means that the point $F$ must also be on a circle with diameter $AC$. So circles with diameters $CD$ and $CA$ intersect at points $C$ and $F$. Consequentially, line $CF$ represent their radical axis and that axis is perpendicular to line $DE$.
In exactly the same way you can show that circles with diameter $CD$ and $BD$ intersect at points $D$ and $G\in BC$. Their radical axis $DG$ is perpendicular to line $CE$.
So you have two radical axis: 
$$CF\perp DE$$
$$DG\perp CE$$
Obviously, radical axis are also heights of triangle $\triangle CDE$. Intersection of these two radical axis defines the orthocenter $H$ of triangle $\triangle CDE$. 
The radical axis theorem states that the three radical axes (for each pair of circles) intersect in one point called the radical center. I won't give the proof here but you can find it on Wikipedia and elsewhere.
So the third radical axis (dotted line) must also pass through point $H$. 
