Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.6 AIME Prove it Ireland

In triangle $ABC$ , the internal and external bisectors of angle $\angle ACB$ meet $AB$ at $D$ and $E$ , respectively. Suppose $B$ is between $A$ and $D$ . If $CD$ is a median of triangle $AEC$ , prove that $|AC| = 3|BC|$ .

Solution

From the Theorem on the Angle Bisector, we know that $D$ and $E$ divide the line segment $AB$ internally and externally in ratio $|AC| : |BC|$ , that is
$\frac{|AC|}{|BC|} = \frac{|AD|}{|BD|} = \frac{|AE|}{|BE|}.$

Since $CD$ is a median of triangle $AEC$ , we have $|AE| = 2|AD|$ . Together with the second equality above this yields
$\frac{|BE|}{|BD|} = \frac{|AE|}{|AD|} = 2,$
hence $|BE| = 2|BD|$ . Therefore, $|AD| = |DE| = |BD| + |BE| = 3|BD|$ and we obtain
$\frac{|AC|}{|BC|} = \frac{|AD|}{|BD|} = 3$
as desired.

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