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

itzz pythogoras theorem

pythagoras

thanq ...

suppose if we take all the angles as x+x+x=180

so x+x=2x which therefore is greater then x

2x>x

easy question.

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!

same question appered in my maths paper........

 umangbansal600, u were a bit wrong. It is AB=EC, not ED. Thus it is AC+EC>AE. Therfore, AC+AB>2AD

 HOW IS IT PYTHOGORAS????????

Anshudude is right!