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Ukrainian National Mathematical Olympiad - Third Round

Geometry Difficulty 5.3 AIME, harder Prove it Ukraine

Let $AD$ be a bisector in an isosceles triangle $ABC$ ( $AB = BC$ ), and let $DE$ be another bisector in the triangle $ABD$ . Find out the measures of all angles in $ABC$ if the bisectors of $ABD$ and $AED$ intersect on the straight line $AD$ .

Solution

Let $K$ be the intersection point for the bisectors of the angles $ABD$ and $AED$ (Fig. 3). Then this point lies on the segment $AD$ and is equidistant from rays $BA$ and $BC$ , as well as from $EA$ and $ED$ . Hence, it's equidistant from rays $DE$ and $DC$ . Then $DA$ is the bisector of $\angle CED$ (in other words, $K$ is an excenter of the triangle $EBD$ ). With the initial conditions, this implies $\angle ADC = 60^\circ$ . Since $\angle DCA = 2\angle DAC$ , we have that $\angle DCA = 80^\circ$ . Finally, we can write $\angle BAC = \angle BCA = 80^\circ$ , $\angle ABC = 20^\circ$ .

Fig. 3

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