Austrian Mathematical Olympiad

We first consider the case where $C$ and $D$ are both not the center of $AB$, so that the tangents in $C$ and $D$ are both not parallel to $AB$.

The tangent in $E$ intersects $AB$ in $X$, the tangent in $F$ intersects $AB$ in $Y$ and the two tangents intersect each other in $Z$. Let $I$ now be the intersection point of the angle bisector of $\angle XYZ$ and $\angle ZXY$. Since the tangent segments $XC$ and $XE$ at $k_1$ are of equal length and $C$, $E$ lie on the legs of the angle $\angle ZXY$, $C$ and $E$ are equidistant from $I$.

The same applies to $D$ and $F$ with the circle $k_2$ and $E$ and $F$ with the semicircle $h$.

This means that the four points lie on a circle with center $I$.

In the remaining special case that $k_1$ passes through the center of $AB$, we can still define $Y$ as the intersection of the tangent in $F$ with $AB$, and $Z$ as the intersection of the tangents in $E$ and $F$. We define $I$ as the intersection of the angle bisectors of $\angle ZYD$ and $\angle EZY$. Therefore, $I$ has the same distance to $DY$ and $YZ$, and the same distance to $EZ$ and $YZ$. This means that $I$ also has the same distance to the parallel lines $DY$ and $EZ$. Thus $I$ lies on the perpendicular bisector of $CE$ and, thus, $IC=IE$.