Fermat 2003 Q17

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Problem Statement

In the diagram, $\triangle ABC$ is right-angled at $C$. If $BD = 2x$, $DC = x$, and $\angle ADC = 2(\angle ABC)$, then the length of $AB$ is

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Solution

$$\text{let } \angle ABC = \theta \implies \angle ADC = 2\theta$$ $$\therefore \angle ADB = 180 - 2\theta$$ $$180 = \angle ABD + \angle ADB + \angle BAD$$ $$180 = \theta + 180 - 2\theta + \angle BAD \implies \angle BAD = \theta$$ $$\therefore \triangle ABD \text{ is isoceles } \implies AD = BD = 2x$$ $$AD^2 = CD^2 + AC^2 \implies AC = x\sqrt{3}$$ $$AB^2 = BC^2 + AC^2 \implies AB^2 = 9x^2 + 3x^2 = 12x^2$$ $$\therefore AB = 2x\sqrt{3}$$

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