<Tri-2025-2247>, hint

Let O be the circumcenter od triangle ABC, then vOB-vOC =vBC=vXB-vXC.

Let X' be the symmetric to X about AH, vXA+vXH=vXX' and AH=XX', [XX']=2R cosA

[BC]=2R sinA. Thus For 2RcosA=2RsinA, <A=45.



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