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Romanian Mathematical Olympiad

Geometry Difficulty 5.2 AIME, harder Prove it Romania

The diagonals of the parallelogram $ABCD$ meet at $O$ . The bisectors of the angles $DAC$ and $DBC$ meet at $T$ . It is known that $TD + TC = TO$ . Find the measures of the angles of triangle $ABT$ .
Claudiu Militaru

Solution

The hypothesis shows that $DOCT$ is a parallelogram. From $AO \parallel DT$ follows $\angle DTA = \angle OAT = \angle DAT$ , hence $DA = DT$ .

This leads to $DA = DT = OC$ ; in the same way $BC = CT = OD$ , therefore $BD = AC$ . So $ABCD$ is a rectangle, $AOTD$ is a rhombus, triangle $AOD$ is equilateral and triangle $ABT$ is equilateral, hence its angles have measures $60^\circ$ .

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