in triangle ABC, E, is any point in the interior of triangle ABC. prove that AB +AC >BE +CE
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Join OA, OB and OC.
According to triangle inequality, sum of any two sides is greater than the third side.
Therefore,
OA+OB > AB ……(1)
OB+OC > BC ……(2)
OC+OA > AC ……(3)
Adding (1), (2) and (3), we get
2(OA+OB+OC) > (AB+BC+AC)
Dividing both sides by 2, we get
(OA+OB+OC) > 1/2(AB+BC+AC).
Shown
According to triangle inequality, sum of any two sides is greater than the third side.
Therefore,
OA+OB > AB ……(1)
OB+OC > BC ……(2)
OC+OA > AC ……(3)
Adding (1), (2) and (3), we get
2(OA+OB+OC) > (AB+BC+AC)
Dividing both sides by 2, we get
(OA+OB+OC) > 1/2(AB+BC+AC).
Shown
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