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Slovenia — National Math Olympiad

Geometry Difficulty 4.6 AIME Prove it Slovenia

In the triangle $ABC$ the bisector of the angle $\angle BAC$ meets the segment $BC$ at $D$ . The triangle $ADC$ is isosceles with the apex at $D$ and the lengths of the segments $CD$ and $BD$ are $|CD| = 36$ and $|BD| = 64$ . Find the lengths of the sides of the triangle $ABC$ .

Solution

Obviously $|BC| = 100$ . Let us find the lengths of the other sides. We have $\angle BAD = \angle DAC = \angle ACB$ , so the triangles $ABD$ and $CBA$ are congruent and $\frac{|AC|}{|AD|} = \frac{|AB|}{|BD|} = \frac{|BC|}{|AB|}$ . The second equality implies $|AB| = \sqrt{|BD|^2 \cdot |BC|} = \sqrt{64 \cdot 100} = 80$ . It then follows from the first equality that $|AC| = \frac{|AD||AB|}{|BD|} = \frac{36 \cdot 80}{64} = 45$ .

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