In a triange ABC, prove that AB + AC > 2AD, where AD is the median from A to BC..pls answer this one....

**Given: **AD is the median of ΔABC.

**To prove: **AB + AC > 2AD

**Construction:** Produce AD to E such that AD = DE. Join CE.

**Proof: **In ΔABD and ΔCDE

AD = DE (Construction)

∠ADB = ∠CDE (Vertically opposite angles)

BD = DC (AD is the median from A to BC)

∴ ΔABD ΔCDE (SAS congruence criterion)

⇒ AB = CE (CPCT) ...(1)

In ΔACE,

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

⇒ AC + AB > AD + DE [Using (1)]

⇒ AC + AB > AD + AD (Constriction)

⇒ AC + AB > 2AD

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