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,(Theorem to be remembered)
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
2.
To prove: AB + BC + CA > 2AD
Construction: AD is joined
Proof: In triangle ABD,
AB + BD > AD [because, the sum of any two sides of a triangle is always greater than the
third side]
----
1
In triangle ADC,
AC + DC > AD [because, the sum of any two
sides of a tri
angle is always greater than the
third side]
----
2
Adding 1 and 2 we get,
AB + BD + AC + DC > AD + AD
=> AB + (BD + DC) + AC > 2AD
=> AB + BC + AC > 2AD
Hence proved
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