Question

prove that the sum of any two sides of a triangle is greater than twice the length of median drawn to the third side

Answer 1

Given: Δ ABC 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)