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Question
Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.
Solution
Given: In triangle ABC with median AD,
To proof: AB + AC > 2AD
AB + BC > 2AD
BC + AC > 2AD
Producing AD to E such that DE = AD and join EC.
Proof: In triangle ADB and triangle EDC,
AD = ED ...[By construction]
∠1 = ∠2 ...[Vertically opposite angles are equal]
DB = DC ...[Given]
So, by SAS criterion of congruence]
ΔADB ≅ ΔEDC
AB = EC ...[CPCT]
And ∠3 = ∠4 ...[CPCT]
Again, in triangle AEC,
AC + CE > AE ...[Sum of the lengths of any two sides of a triangle must be greater than the third side]
AC + CE > AD + DE
AC + CE > AD + AD ...[AD = DE]
AC + CE > 2AD
AC + AB > 2AD ...[Because AB = CE]
Hence proved.
Similarly, AB + BC > 2AD and BC + AC > 2AD.
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