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

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

Answers (1)

 

Given: ABC is a triangle

AD is the median at BC drawn from A.

Produce D to E and DE = AD, Join DE

To Prove: AB + AC > AD

Proof: Let \triangleABD and \triangleECD

CD = DB                                 (D is mid-point)

\angleBDA = \angleCDE          (Vertically opposite angle)

ED = AD                                 (by construction)

\therefore \triangle ABD \cong \triangleECD       (by SAS congruence)

Þ EC = AB                 (by CPCT)

In \triangleACE, AC + EC > AE                    (Sum of two sides is greater than the third side)

Then AC + AB > AD + DE     (QAB = EC and AE = AD + DE as D is the mid-point)

AC + AB > 2AD

AC + AB > AD

Hence proved.

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