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Junior Balkan Mathematical Olympiad Shortlist

Geometry Difficulty 5.5 AIME, harder Prove it JBMO

Problem:
Let $ABC$ be a triangle in which $BL$ is the angle bisector of $\widehat{ABC}$ ( $L \in AC$ ), $AH$ is an altitude of $\triangle ABC$ ( $H \in BC$ ), and $M$ is the midpoint of the side $[AB]$ . It is known that the midpoints of the segments $[BL]$ and $[MH]$ coincide. Determine the internal angles of triangle $\triangle ABC$ .

Solution

Solution:
Let $N$ be the intersection of the segments $[BL]$ and $[MH]$ . Because $N$ is the midpoint of both segments $[BL]$ and $[MH]$ , it follows that $BMLH$ is a parallelogram. This implies that $ML \parallel BC$ and $LH \parallel AB$ and hence, since $M$ is the midpoint of $[AB]$ , the angle bisector $BL$ and the altitude $AH$ are also medians of $\triangle ABC$ . This shows that $\triangle ABC$ is an equilateral one with all internal angles measuring $60^{\circ}$ .

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