Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Prove it Ireland

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
$\frac{|AB|}{|PF|} = \sqrt{2}.$

Solution

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

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

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

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