Prove that the sum of any two sides of a tria | vedclass

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(N/A) Given: $	riangle ABC$ with median $AD$ to side $BC$.To prove: $AB + AC > 2AD$.Construction: Produce $AD$ to $E$ such that $DE = AD$ and join $E

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(N/A) Given: $	riangle ABC$ with median $AD$ to side $BC$.To prove: $AB + AC > 2AD$.Construction: Produce $AD$ to $E$ such that $DE = AD$ and join $EC$.Proof: In $	riangle ADB$ and $	riangle EDC$:$AD = ED$ (By construction)$\angle 1 = \angle 2$ (Vertically opposite angles)$DB = DC$ ($AD$ is the median)By $SAS$ congruence criterion,$	riangle ADB \cong 	riangle EDC$.Therefore,$AB = EC$ $(CPCT)$.Now,in $	riangle AEC$,the sum of any two sides is greater than the third side:$AC + CE > AE$Since $AE = AD + DE = AD + AD = 2AD$ and $CE = AB$:$AC + AB > 2AD$.Hence,it is proved that the sum of two sides is greater than twice the median to the third side.

Prove that the sum of any two sides of a triangle is greater than twice the median with respect to the third side.

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