Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Singapore

In $\triangle ABC$ , $AB = AC = 14\sqrt{2}$ , $D$ is the midpoint of $CA$ and $E$ is the midpoint of $BD$ . Suppose $\triangle CDE$ is similar to $\triangle ABC$ . Find the length of $BD$ .

Solution

Let $\ell = AB = AC = 14\sqrt{2}$ and $BC = x$ . The 4 angles marked in the figure are all equal. This implies that $\triangle ABC$ , $\triangle BCD$ , $\triangle CDE$ are all similar. Thus $BD = BC = x$ .

Also $AB/BC = BC/CD$ . That is $\ell/x = x/(\ell/2)$ . From this we get $x = \ell/\sqrt{2}$ . Since $\ell = 14\sqrt{2}$ , we have $x = 14$ .

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