Here is an alternative way to show that $EG\perp HF$. 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,
$$\angle CQD=\pi-\gamma-\delta\text{ so }\angle SQB=\frac{\angle CQD}{2}=\frac{\pi}{2}-\frac{\gamma}{2}-\frac{\delta}{2}\,.$$
Similarly,
$$\angle AQD=\pi-\alpha-\delta\text{ so }\angle SPB=\frac{\angle CQD}{2}=\frac{\pi}{2}-\frac{\alpha}{2}-\frac{\delta}{2}\,.$$
The sum of internal angles of the quadrilateral $PSQB$ is $2\pi$, whence
$$\begin{align}\angle PSQ&=2\pi-\big(\angle SQB+\angle SPB+(2\pi-\angle PBQ)\big)\\&=\angle PBQ - \angle SQB-\angle SPB\,.\end{align}$$
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 $\alpha+\beta+\gamma+\delta=2\pi$, 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 $EG\perp HF$. The rest goes as Marco's answer.