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

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!

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