IRL ABooklet

Geometry Difficulty 4.5 AIME Prove it Ireland

The points $A$ , $B$ , $P$ , $Q$ are collinear, and $ABCD$ is a parallelogram. The lines $PD$ and $BC$ meet at $E$ . The lines $QC$ and $AD$ meet at $F$ . The lines $PF$ and $BC$ meet at $H$ . The lines $QE$ and $AD$ meet at $G$ .
Prove that $GH$ is parallel to $AB$ .

Solution

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.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.