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Ukrainian National Mathematical Olympiad, 3rd Round

Geometry Difficulty 4.7 AIME Prove it Ukraine

In triangle $ABC$ $AL$ , $BM$ , $CN$ are medians. Prove, that $\angle ANC = \angle ALB$ if and only if $\angle ABM = \angle LAC$ .

Solution

Lines $LN$ and $AC$ are parallel, hence $\angle NLA = \angle LAC$ . We have to show the following implication (fig. 7):
$\angle ANC = \angle ALB \Leftrightarrow \angle ABM = \angle NLA.$
Let $G$ be a centroid, then we have the following equivalences. $\angle ANC = \angle ALB \Leftrightarrow \angle BNC + \angle ALB = \pi \Leftrightarrow BNLG$ - cyclic $\Leftrightarrow \angle ABM = \angle NLA$ , since they share the common segment. The statement is proved

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.