IRL_ABooklet_2023

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



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

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