IRL_ABooklet_2023

From the construction we have $|EF|=|FD|$ and $FP$ is parallel to $AB$. It follows that $△ABE$ is similar to $△PFE$, hence $$\frac{|AB|}{|PF|}=\frac{|BE|}{|EF|}(2)$$

Because $BC\parallel AD$, the angles $\angle FBP$ and $\angle BDA$ are equal. This implies that the right angled triangles $△ABD$ and $△PFB$ are similar, and so $$\frac{|AB|}{|PF|}=\frac{|BD|}{|BF|}=\frac{|BE|+2\cdot |EF|}{|BE|+|EF|}(3)$$

Combining (2) and (3) we get $$\begin{array}{c}\frac{|BE|}{|EF|}=\frac{|BE|+2\cdot |EF|}{|BE|+|EF|},\text{ i.e.}\\ |BE{|}^2+|BE|\cdot |EF|=|BE|\cdot |EF|+2\cdot |EF{|}^2\text{which gives}\\ |BE{|}^2=2\cdot |EF{|}^2.\end{array}$$ Using (2) we now obtain $$\frac{|AB|}{|PF|}=\frac{|BE|}{|EF|}=\sqrt{2}.$$