Question

Prove that any two sides of a $\Delta$ are together greater than twice the median drawn to the third side.

Answer 1

Given:Triangle ABC in which AD is the median.

To prove: $AB+AC>2AD$

Proof:In $\Delta ADB$ and $\Delta EDC$

$AD=DE$ [By construction]

D is the midpoint BC.[$DB=DB$]

$\Delta ADB=\Delta EDC$ [vertically opposite angles]

Therefore $\Delta ADB\cong \Delta EDC$ [By SAS congruence criterion.]

$AB=ED$ [Corresponding parts of congruent triangles]

In$\Delta AEC$, $AC+ED>AE$[sum of any two sides of a triangle is greater than the third side]

$AC+AB>2AD$ [$AE=AD+DE=AD+AD=2AD$ and $ED=AB$]