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Geometry Difficulty 4.7 AIME Prove it Ukraine

In triangle ABCABC ALAL, BMBM, CNCN are medians. Prove, that ∠ANC=∠ALB\angle ANC = \angle ALB if and only if ∠ABM=∠LAC\angle ABM = \angle LAC.

Solution

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

Figure 1

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