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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we know that from any point perpendicular is the shortest that means

AB>AD eq. 1 and also ,AC>AD eq.  2

adding 1 and  2

AB+AC>AD+AD

AB+AC>2AD

hope it helps ....... cheers.......)

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