Prove that Sum of Two Sides of a triangle is greater than twice the length of median drawn to third side.

Dear Student!

Given: ΔABC in which AD is a median.

To prove: AB + AC > 2AD.

Construction: Produce AD to E, such that AD = DE. Join EC.

Proof: In ΔADB and ΔEDC,

AD = DE              (Construction)

BD = BD             (D is the mid point of BC)

∠ADB = ∠EDC       (Vertically opposite angles)

∴ ΔADB ΔEDC   (SAS congruence criterion)

⇒ AB = ED               (CPCT)

In ΔAEC,

AC + ED > AE           (Sum of any two sides of a triangles is greater than the third side)

∴ AC + AB > 2AD      (AE = AD + DE = AD + AD = 2AD & ED = AB)

Cheers!