question_answer

If O is a point in the exterior of \[\Delta ABC\], show that: 2 (OA + OB + OC) > (AB + BC + CA).

Answer:

In\[\Delta AOB\], OA + OB > AB             ....(i) Similarly, in ABOC, OB + OC > BC                         ....(ii) and in\[\Delta AOC\], OA + OC > AC             ....(iii)                                                                                                    By adding (i), (ii) and (iii), we get OA + OB + OB + OC + OA + OC > AB + BC +CA 20A + 20B + 20C > AB + BC + CA 2(OA+OB+OC)>AB+BC+CA