The Intercept Theorem applied to $AG \parallel BH$ and lines intersecting at $P$ gives
$$\frac{|PA|}{|PB|} = \frac{|AF|}{|BH|} \quad \text{and} \quad \frac{|PA|}{|PB|} = \frac{|AD|}{|BE|}, \text{ hence } |AF| \cdot |BE| = |BH| \cdot |AD|.$$

Considering lines that intersect at $Q$, we obtain in a similar way
$$\frac{|QA|}{|QB|} = \frac{|AG|}{|BE|} \quad \text{and} \quad \frac{|QA|}{|QB|} = \frac{|AF|}{|BC|}, \text{ hence } |AF| \cdot |BE| = |AG| \cdot |BC|.$$

Taking into account $|AD| = |BC|$, these two equations imply $|BH| = |AG|$. Since $AG$ and $BH$ are parallel, this implies that $ABHG$ is a parallelogram.