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)
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