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Singapore Mathematical Olympiad (SMO)

Geometry Difficulty 4.9 AIME Prove it Singapore

In $\triangle ABC$ , $AB = AC$ , $D$ is a point on the side $BC$ and $E$ is a point on the segment $AD$ . Given that $\angle BED = \angle BAC = 2\angle CED$ , prove that $BD = 2CD$ .

Solution

Let $\angle CED = x$ and $\angle ABE = y$ . Then
$\angle BAC = \angle BED = 2x \Rightarrow \angle BAE = 2x - y \text{ and } \angle EAC = y.$
Let $F$ be the point on $BE$ so that $\angle AFE = \pi$ . (Note that $x = \angle CED = y + \angle ECA$ implying $x > y$ . Thus $F$ is in fact in the interior of the segment $BE$ .) Since $\angle AFE = x$ and $\angle FED = 2x$ , we have $AE = EF$ . Next we have
$AB = AC \Rightarrow \triangle BAF \cong \triangle ACE \ (\text{ASA}) \Rightarrow BF = AE = EF$
∴ $[AEC] = [BFA] = [AFE] \Rightarrow [BEA] = 2[AEC] \Rightarrow BD = 2CD$ .

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