Question

Prove that any to side of a triangle are together greater than twice the median draw to the third side.
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Answer 1

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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 ADC and EDB,

AD = DE (by const)

DC = BD (as D is mid-point)

ADC = EDB (vertically opp. s)

Therefore,

In ABE, ADC 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.

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