Advanced Geometric Problem & Solution (Twice a week update) Now 4620 problems posted.
<Tri-2025-2209>, hint
Since circumcircles of triangles ABC, HBC are congruent
MO cuts BC in midpoint D of BC ans so MD=DO = R cosA
AH=2RcosA. AH//MO. Hence AOMH is a parallelogram , so AM cuts in the midpoint of OH.
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