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

Step:Proof

Given: $\Delta ABC$ with median $AD$.

To Prove: $AB+AC>2AD,AB+BC>2AD,BC+AC>2AD$

Construction:

Extend $AD$ to $E$ such that $DE=AD$,Join $EC$.

Proof: From $\Delta ADB$ and $\Delta EDC,$

$AD=ED$ [By construction]

$\angle 1=\angle 2$ [Vertically opposite angles are equal]

$DB=DC$ [Given]

By$SAS$ criterion of congruence,($SAS$ congruence rule: If any two sides and the angle included between the sides of one triangle are equivalent to the corresponding two sides and the angle between the sides of the second triangle, then the two triangles are said to be congruent by $SAS$ rule.)

$\Delta ADB\cong \Delta EDC$

$AB=EC\left[CPCT\right]$

And $\angle 3=\angle 4\left[CPCT\right]$

Now, in$\Delta AEC,$

Since sum of the lengths of any two sides of a triangle must be greater than the third side,We have

$AC+CE>AE$

$\Rightarrow AC+CE>AD+DE$

$\Rightarrow AC+CE>AD+AD[AD=DE]$

$\Rightarrow AC+CE>2AD$

$\Rightarrow AC+AB>2AD[\because AB=CE]$

Similarly,We get,

$AB+BC>2ADandBC+AC>2AD.$

Hence, proved.