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Geometry Difficulty 4.9 AIME Prove it United States

Problem:

Consider triangle $ABC$ with $\angle A = 2 \angle B$ . The angle bisectors from $A$ and $C$ intersect at $D$ , and the angle bisector from $C$ intersects $\overline{AB}$ at $E$ . If $\frac{DE}{DC} = \frac{1}{3}$ , compute $\frac{AB}{AC}$ .

Solution

Solution:

$\quad \frac{7}{9} \quad$ Let $AE = x$ and $BE = y$ . Using angle-bisector theorem on $\triangle ACE$ we have $x : DE = AC : DC$ , so $AC = 3x$ . Using some angle chasing, it is simple to see that $\angle ADE = \angle AED$ , so $AD = AE = x$ . Then, note that $\triangle CDA \sim \triangle CEB$ , so $y : (DC + DE) = x : DC$ , so $y : x = 1 + \frac{1}{3} = \frac{4}{3}$ , so $AB = x + \frac{4}{3}x = \frac{7}{3}x$ . Thus the desired answer is $AB : AC = \frac{7}{3}x : 3x = \frac{7}{9}$ .

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