IRL ABooklet

Geometry Difficulty 4.5 AIME Prove it Ireland

Through a point P on the hypotenuse $AB$ of the right-angled triangle $ABC$ lines are drawn parallel to the other two sides. These parallels meet $BC$ and $AC$ at $D$ and $E$ , respectively. Prove $|BP| \cdot |AP| = |BD| \cdot |CD| + |AE| \cdot |CE|$ .

Solution

Because PE is parallel to $BC$ and PD is parallel to $AC$ , the two right angled triangles $APE$ and $PBD$ are similar. This implies that there exists a positive number $s$ (the similarity factor) such that
$|AP| = s|BP|, \quad |AE| = s|DP| \quad \text{and} \quad |EP| = s|BD|.$

Using these equations together with $|DP| = |CE|$ and $|EP| = |CD|$ , which are true since $CEPD$ is a rectangle, the Theorem of Pythagoras for triangle $PBD$ implies the desired identity:
$\begin{aligned} s|BP|^2 &= s|BD|^2 + s|DP|^2 \\ |BP| \cdot |AP| &= |BD| \cdot |CD| + |AE| \cdot |CE|. \end{aligned}$

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