Question

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

Answer 1

Given$:$ In $△ABC,AD$ is the median drawn from $A$ to $BC$

To Prove that $AB+AC>AD$

Construction$:$Produce $AD$ to $E$ so that $DE=AD$, join $BE$

Proof$:$In $△ADC$ and $△EDB$ we have

$AD=DE$(constant)

$DC=BD$ as $D$ is the midpoint

$\angle ADC=\angle EDB$ (vertically opposite angles)

$\therefore$ In $△ABE,$ $△ADC\cong △EDB$ by S.A.S

This gives $BE=AC$

$AB+BE>AE$

$AB+AC>2AD$ $\because AD=DE$ and $BE=AC$

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