Advanced Geometric Problem & Solution (Twice a week update) Now 4610 problems posted.
<Tri-2025-2228>, hint
Let E on the line AB such that BE=BC
Since <B=2<C, <BEC=<BCE=<C .Triangles AEC and ACB are similar
AC/AE=AC/AB+BC=AB/AC.
In triangle AEB , AC/sinC=(AB+BC)/sin2C=(AB+BC)/2sinCcosC
2AC cosC=(AB+BC) ---> 2AC>AB+BC
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