Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Prove it Croatia

Let $ABC$ be a right triangle with the right angle at vertex $C$ . Let $D$ be a point on the segment $AC$ and let $E$ be a point on the segment $BD$ such that $\angle ABC = \angle DAE = \angle AED$ . Prove that $|BE| = 2|CD|$ . (Lithuania 2010)

Solution

Let $\triangle ABC = \triangle DAE = \triangle AED = \beta$ .

Then
$\begin{aligned} \angle BAC &= 90^\circ - \beta, \\ \angle BAE &= (90^\circ - \beta) - \beta = 90^\circ - 2\beta, \\ \angle BEA &= 180^\circ - \beta. \end{aligned}$
Applying the law of sines to the triangle $ABE$ :
$|BE| = \frac{\sin \angle BAE}{\sin \angle AEB} \cdot |AB| = \frac{\sin(90^\circ - 2\beta)}{\sin(180^\circ - \beta)} \cdot |AB| = \frac{\cos(2\beta)}{\sin \beta} \cdot |AB|.$
Since $|CD| = |BC| \operatorname{ctg} \angle CDB = |BC| \operatorname{ctg}(2\beta)$ , we finally obtain
$\frac{|BE|}{|CD|} = \frac{\frac{\cos(2\beta)}{\sin \beta} \cdot |AB|}{|BC| \operatorname{ctg}(2\beta)} = \frac{\cos(2\beta)|AB|}{\sin \beta \cdot |BC| \cdot \frac{\cos(2\beta)}{\sin(2\beta)}} = \frac{\sin(2\beta)|AB|}{\sin \beta \cdot |AB| \cos \beta} = 2.$

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