Question 125
A parallelogram *ABCD* is such that $BD \perp AB$. A point $P$ on $BC$ is such that $|PD| = |BE|$ where $E$ is the intersection point of $BD$ and $AP$. The point $F$ is the foot of the perpendicular from $P$ to $BD$. Prove
$$
\frac{|AB|}{|PF|} = \sqrt{2}.
$$
$$
\frac{|AB|}{|PF|} = \sqrt{2}.
$$