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Ukrainian National Mathematical Olympiad, Third Round, First Tour

Geometry Difficulty 4.8 AIME Prove it Ukraine

Let $AL$ be a bisector of triangle $ABC$ . The circle centered at $B$ with radius $BL$ meets the ray $AL$ at point $E$ , and the circle centered at $C$ with radius $CL$ meets the ray $AL$ at point $D$ (points $E$ and $D$ are different from point $L$ ). Prove that $AL^2 = AE \cdot AD$ .
(Mykola Moroz)

Solution

Clearly, triangles $BEL$ and $CDL$ are isosceles, and angles $\angle CLD$ and $\angle BLE$ are vertical (fig. 3). Then $\angle BEL = \angle BLE = \angle CLD = \angle CDL$ . Then $\angle AEB = \angle ALC$ as adjacent to equal angles. Also $\angle CAL = \angle BAL$ , as $AL$ is a bisector.

Note that triangles $CLA$ and $BEA$ are similar by two angles. Then $\frac{AL}{AE} = \frac{AC}{AB}$ , from where $AL = AE \cdot \frac{AC}{AB}$ . Also note that triangles $BLA$ and $ACD$ are similar by two angles. Then $\frac{AL}{AD} =$
$\frac{AB}{AC}$

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