If O is the point in the exterior of∆ABC, show that
2(OA+OB+OC)>(AB+BC+CA).
Answers
ANSWER:
the sum of two side is always greater than the third side.
the sum of two side is always greater than the third side.In triangle ABC,
the sum of two side is always greater than the third side.In triangle ABC,O is the point that lies in the EXTERIOR of the triangle ABC.
triangle OAB , triangle OAC and triangle BOC are formed exterior of triangle ABC.
the sum of:
the sum of:OA+OB > AB
the sum of:OA+OB > ABOA+OC > AC
the sum of:OA+OB > ABOA+OC > ACOB+OC > BC
the sum of:OA+OB > ABOA+OC > ACOB+OC > BCadd....
the sum of:OA+OB > ABOA+OC > ACOB+OC > BCadd....OA+OB+OA+OC+OB+OC > AB+AC+BC
the sum of:OA+OB > ABOA+OC > ACOB+OC > BCadd....OA+OB+OA+OC+OB+OC > AB+AC+BC2OA+2OB+2OC > AB+AC+BC
the sum of:OA+OB > ABOA+OC > ACOB+OC > BCadd....OA+OB+OA+OC+OB+OC > AB+AC+BC2OA+2OB+2OC > AB+AC+BC2(OA+OB+OC) > PERIMETER OF ∆ABC