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Ukrainian National Mathematical Olympiad - Third Round (Second Tour)

Geometry Difficulty 5.0 AIME Prove it Ukraine

An acute-angled triangle $ABC$ has the side $BC > AB$ , and the bisectrix $BL = AB$ . On the segment $BL$ there exists point $M$ , for which $\angle AML = \angle BCA$ . Prove that $AM = LC$ .

Fig. 16

Solution

On the segment $BC$ put a point such as $BD = BL$ (fig. 16). Then $\Delta ABL = \Delta BLD$ , mark $\angle LAB = \angle BLA = \angle BLD = \angle BDL$ . On the segment $BD$ choose a point $K$ , such that $\angle ALM = \angle DLK$ . Then $\Delta ALM = \Delta DLK$ because of $LD = AL$ , but then $\angle AML = \angle LKD = \angle BCA$ , that's why $KL = AM = LC$ , and it is exactly what we have to prove.

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