prove that the sum of the side of a triangle is greater than the sum of its three median
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Let AD,BE & CF be the three medians of a ∆ABC.
WE KNOW THAT THE SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER THAN TWICE THE MEDIAN DRAWN TO THE THIRD SIDE.
AB+AC>2AD;. AB+BC>2BE & BC+AC>2CF.
Adding We get,
2(AB+BC+AC) >2(AD+BE+CF)
(AB+BC+AC) >(AD+BE+CF)
Hence, the perimeter of a triangle is greater than the sum of its three medians.
Let AD,BE & CF be the three medians of a ∆ABC.
WE KNOW THAT THE SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER THAN TWICE THE MEDIAN DRAWN TO THE THIRD SIDE.
AB+AC>2AD;. AB+BC>2BE & BC+AC>2CF.
Adding We get,
2(AB+BC+AC) >2(AD+BE+CF)
(AB+BC+AC) >(AD+BE+CF)
Hence, the perimeter of a triangle is greater than the sum of its three medians.
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