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Bulgarian Spring Tournament

Geometry Difficulty 4.5 AIME Prove it Bulgaria

Given is a triangle ABCABC and two points D∈ACD \in AC, E∈BDE \in BD such that ∠DAE=∠AED=∠ABC\angle DAE = \angle AED = \angle ABC. Show that BE=2CDBE = 2CD iff ∠ACB=90∘\angle ACB = 90^\circ.
(Nikolay Nikolov)

Solution

Let α=∠A\alpha = \angle A, β=∠B\beta = \angle B and γ=∠C\gamma = \angle C. Then
BEsin⁡(α−β)=ABsin⁡(π−β),CDsin⁡(α−β)=BCsin⁡2β,ABsin⁡γ=BCsin⁡α \frac{BE}{\sin(\alpha - \beta)} = \frac{AB}{\sin(\pi - \beta)}, \quad \frac{CD}{\sin(\alpha - \beta)} = \frac{BC}{\sin 2\beta}, \quad \frac{AB}{\sin \gamma} = \frac{BC}{\sin \alpha}
and so
BECD=sin⁡2βsin⁡γsin⁡βsin⁡α=2cos⁡βsin⁡γsin⁡α=2cos⁡βsin⁡γsin⁡βcos⁡γ+cos⁡βsin⁡γ \frac{BE}{CD} = \frac{\sin 2\beta \sin \gamma}{\sin \beta \sin \alpha} = \frac{2 \cos \beta \sin \gamma}{\sin \alpha} = \frac{2 \cos \beta \sin \gamma}{\sin \beta \cos \gamma + \cos \beta \sin \gamma}
Therefore BE=2CD⇔cos⁡γ=0⇔γ=90∘BE = 2CD \Leftrightarrow \cos \gamma = 0 \Leftrightarrow \gamma = 90^\circ. □

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