Prove that the perimeter of a triangle is greater than the sum of its 3 medians
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Let ABC be the triangle and D. E and F are midpoints of BC, CA and AB respectively.
Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side, Hence in ΔABD, AD is a median
AB + AC > 2AD. (i)
Similarly, we get
BC + AC > 2CF . (ii)
BC + AB > 2BE. (iii)
On adding the equations (i),(ii) and (iii), we get
(AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE
2(AB + BC + AC) > 2(AD + BE + CF)
∴ AB + BC + AC > AD + BE + CF
Thus the perimeter of a triangle is greater than the sum of the medians
Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side, Hence in ΔABD, AD is a median
AB + AC > 2AD. (i)
Similarly, we get
BC + AC > 2CF . (ii)
BC + AB > 2BE. (iii)
On adding the equations (i),(ii) and (iii), we get
(AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE
2(AB + BC + AC) > 2(AD + BE + CF)
∴ AB + BC + AC > AD + BE + CF
Thus the perimeter of a triangle is greater than the sum of the medians
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