Prove that the perimeter 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 in equations,
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 in equations,
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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