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Prove that the sum of any sides of a triangle is greater than twice the length of median drawn to the third side.?
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Prove that the sum of any sides of a triangle is greater than twice th...


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)
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