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Geometry Difficulty 4.9 AIME Prove it United States

Problem:

In $\triangle ABC$ , the external angle bisector of $\angle BAC$ intersects line $BC$ at $D$ . $E$ is a point on ray $\overrightarrow{AC}$ such that $\angle BDE = 2 \angle ADB$ . If $AB = 10$ , $AC = 12$ , and $CE = 33$ , compute $\frac{DB}{DE}$ .

Solution

Solution:

Let $F$ be a point on ray $\overrightarrow{CA}$ such that $\angle ADF = \angle ADB$ . $\triangle ADF$ and $\triangle ADB$ are congruent, so $AF = 10$ and $DF = DB$ . So, $CF = CA + AF = 22$ . Since $\angle FDC = 2 \angle ADB = \angle EDC$ , by the angle bisector theorem we compute $\frac{DF}{DE} = \frac{CF}{CE} = \frac{22}{33} = \frac{2}{3}$ .

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