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