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Ukrainian National Mathematical Olympiad, 3rd Round

Geometry Difficulty 4.9 AIME Prove it Ukraine

Triangle ABC is inscribed into the circle. Tangents at points A and B meet at point T. The line, that passes through the point T and is parallel to AC, intersects BC at point D. Prove that $AD = CD$ .

Solution

Since triangle $ATB$ is isosceles, $\angle TBA$ and $\angle BCA$ share common arc, then:

$\angle TBA = \angle TAB = \angle BCA = \angle BDT.$

This implies, that $BTAD$ is cyclic (fig. 9). Hence $\angle BTA = \angle ADC$ and triangles $ABT$ and $ACD$ are similar. Triangle $ATB$ is isosceles, thus so is $\square ACD$ . This implies that $AD = CD$ .

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