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Slovenia — National Math Olympiad, 2012

Geometry Difficulty 4.4 AIME Prove it Slovenia

Let $ABC$ be a triangle. Let $D$ denote the intersection point of the bisector of the angle $\angle BAC$ and the side $BC$ , and let $E$ denote the intersection point of the bisector of the angle $\angle CBA$ and the side $AC$ . Suppose $|CD| = |CE|$ . Prove that the triangle $ABC$ is isosceles.

Solution

Denote by $F$ the intersection point of the lines $AD$ and $BE$ . Then the line $CF$ is the bisector of the angle $\angle ACB$ . The triangles $CFD$ and $CFE$ coincide in two sides and the angle between them, hence they are congruent. The angles $\angle CDF$ and $\angle FEC$ are thus equal. The triangles $ADC$ and $BEC$ coincide in one side and the adjacent angles, so they are congruent. Hence the angles $\angle DAC$ and $\angle CBE$ are equal. From this we may obviously conclude that the triangle $ABC$ is isosceles with the top angle at $C$ .

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