Csdrhrt

Let $ABCD$ be a cyclic quadrilateral whose opposite sides are not parallel. The lines $AB$ and $CD$ intersect at point $P$. The lines $AD$ and $BC$ intersect in point $Q$. The bisector of the angle $angle DPA$ cuts the line segment $BC$ and $DA$ in the points $E$ and $G$, respectively. The bisector of the angle $angle AQB$ cuts the line segments $AB$ and $CD$ in the points $H$ and $F$.

Now it seems as if the quadrilateral $EFGH$ is a always a rhombus. I intend to prove this.

Maybe anyone has a checklist or any idea to begin with.

Here is an alternative way to show that $EGperp HF$. In fact, I shall verify that, for any convex quadrilateral $ABCD$, the quadrilateral $EFGH$ is a rhombus if and only if the quadrilateral $ABCD$ is cyclic. Without loss of generality, suppose that the configuration of points $P$ and $Q$ are as in the OP's figure (that is, $P$ and the segment $AD$ are on the opposite side of the line $BC$, and $Q$ and the segment $CD$ are on the opposite side of the line $AB$).

Let $EG$ and $FH$ meet at $S$. Write $alpha$, $beta$, $gamma$, and $delta$ for the angles $angle DAB$, $angle ABC$, $angle BCD$, and $angle CDA$, respectively. Then,
$$angle CQD=pi-gamma-deltatext{ so }angle SQB=frac{angle CQD}{2}=frac{pi}{2}-frac{gamma}{2}-frac{delta}{2},.$$
Similarly,
$$angle AQD=pi-alpha-deltatext{ so }angle SPB=frac{angle CQD}{2}=frac{pi}{2}-frac{alpha}{2}-frac{delta}{2},.$$
The sum of internal angles of the quadrilateral $PSQB$ is $2pi$, whence
$$begin{align}angle PSQ&=2pi-big(angle SQB+angle SPB+(2pi-angle PBQ)big)\&=angle PBQ - angle SQB-angle SPB,.end{align}$$
However, $angle PBQ=angle ABC=beta$, so
$$angle PSQ=beta-left(frac{pi}{2}-frac{gamma}{2}-frac{delta}{2}right)-left(frac{pi}{2}-frac{alpha}{2}-frac{delta}{2}right),.$$
That is,
$$angle PSQ=(beta+delta)+frac{alpha+gamma}{2}-pi=frac{beta+delta}{2}+frac{alpha+beta+gamma+delta}{2}-pi,.$$
Since $alpha+beta+gamma+delta=2pi$, we have
$$angle PSQ=frac{beta+delta}{2},.$$

If $EGHF$ is a rhombus, then $angle PSQ=dfrac{pi}{2}$, making $beta+delta=pi$. Consequently, the quadrilateral $ABCD$ is cyclic. Conversely, if the quadrilateral $ABCD$ is cyclic, then $beta+delta=pi$ implies that $angle PSQ=dfrac{pi}{2}$, so $EGperp HF$. The rest goes as Marco's answer.

We first show that $EG perp HF$. Let $E'$ and $G'$ be the intersections of line $GE$ with the circle so that $G$ is between $G'$ and $E$. Similarly, define $H'$ and $F'$. For any two points $X,Y$ on the circle, let $XY$ denote the radian measure of the shorter arc connecting them (sorry, I couldn't figure out the arc command here). By the assumptions, we have
$$begin{align}newcommand{arc}[1]{overset{mmlToken{mo}{⏜}}{#1}}arc{H'E'}+arc{G'F'}&=arc{H'B}+arc{BE'}+arc{G'D}+arc{DF'}=(arc{H'B}+arc{DF'})+(arc{BE'}+arc{G'D})\&=(arc{CF'}+arc{AH'})+(arc{E'C}+arc{AG'})=arc{E'C}+arc{CF'}+arc{H'A}+arc{AG'}\&=arc{E'F'}+arc{H'G'},end{align}$$
which implies our claim.

Let $S$ be the intersection of $EG$ and $HF$.
Now, in triangle $triangle PHF$ the angle bisector of $angle P$ is perpendicular ot $HF$, hence it is an isosceles triangle, hence $S$ is the midpoint of the side $HF$. Similarly, $S$ is the midpoint of side $EG$. So the quadrilateral $EFGH$ has perpendicular diagonals that bisect each other. This happens only if $EFGH$ is a rhombus.