prove that the sum of the three sides of a triangle is greater than the sum of its three 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 > 2(AD)
Similarly, we get
BC + AC > 2CF
BC + AB > 2BE
On adding the above inequations, 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
Hence, we can say that 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 > 2(AD)
Similarly, we get
BC + AC > 2CF
BC + AB > 2BE
On adding the above inequations, 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
Hence, we can say that the perimeter of a triangle is greater than the sum of the medians.
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