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Question
Prove that the perimeter of a triangle is greater than the sum of its three medians.
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Solution
Given: ΔABC is on which AD, BE and CF are its medians.
To Prove: We know that the sum of any two sides of a triangle is greater than twice the median bisecting the third side.
Therefore, AD is the median bisecting BC
⇒ AB + AC > 2AD ...(i)
BE is the median bisecting AC ...(ii)
And CF is the median bisecting AB
⇒ BC + AC > 2EF ...(iii)
Adding (i), (ii) and (iii), we get
(AB + AC) + (AB + BC) + (BC + AC) > 2. AD + 2 . BE + 2 . BE + 2 . CF
⇒ 2 (AB + BC + AC) > 2 (AD + BE + CF)
⇒ AB + BC + AC > AD + BE + CF.
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