Question

Sum of any two sides of a triangle is greater than twice the median proved

Answer 1

let ABC is a triangle and AD be the median.

In ∆ABD,

AB+BD>AD.......eq(1),

now, in ∆ACD

AC+CD>AD,.......eq(2),

adding these two equations, we get

AB+BD+AC+CD>AD+AD,

AB+BC+CD>2.AD

Answer 2

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Sum of any two sides of a triangle is greater than twice the median

________________________

Given: In triangle ABC, AD is the median drawn from A to BC.

To prove: AB + AC > AD

Construction: Produce AD to E so that DE = AD, Join BE.

Proof:

Now in
Δ ADC andΔ EDB,

AD = DE (by const)

DC = BD(as D is mid-point)

Δ ADC =Δ EDB (vertically opp.s)

Therefore,

InΔ ABEΔ ,ADC ≅ Δ EDB(by SAS)

This gives, BE = AC.

AB + BE > AE

AB + AC > 2AD (AD = DE and BE = AC)

Hence the sum of any two sides of a triangle is greater than the median drawn to the third side.