<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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