Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Prove it JBMO

Problem:
Let ABCABC be a triangle in which BLBL is the angle bisector of ABC^\widehat{ABC} (LACL \in AC), AHAH is an altitude of ABC\triangle ABC (HBCH \in BC), and MM is the midpoint of the side [AB][AB]. It is known that the midpoints of the segments [BL][BL] and [MH][MH] coincide. Determine the internal angles of triangle ABC\triangle ABC.

Solution

Solution:
Let NN be the intersection of the segments [BL][BL] and [MH][MH]. Because NN is the midpoint of both segments [BL][BL] and [MH][MH], it follows that BMLHBMLH is a parallelogram. This implies that MLBCML \parallel BC and LHABLH \parallel AB and hence, since MM is the midpoint of [AB][AB], the angle bisector BLBL and the altitude AHAH are also medians of ABC\triangle ABC. This shows that ABC\triangle ABC is an equilateral one with all internal angles measuring 6060^{\circ}.

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