Question

prove that in any triangle the sum of three sides of a triangle is greater than twice any one of the median​

Answer 1

   

See the fig. I'm considering ΔABC. CD is a median. We have to prove that AC + BC + AB > 2CD.

CD is produced towards E such that CD = DE & CE = 2CD, and AE is joined.

Consider triangles ADE and BDC.

CD = DE (As constructed)

AD = BD (∵ CD is the median to AB so that D is its midpoint)

∠ADE = ∠BDC (Alternate angles)

∴ According to SAS theorem, ΔADE ≅ ΔBDC.

∴ AE = BC.

We know that the largest side of any triangle is always less than the sum of the other two. So in ΔACE,

⇒ AC + AE > CE

⇒ AC + BC > 2CD  [∵ AE = BC  &  CE = 2CD]

⇒ AC + BC + AB > 2CD + AB

So that  AC + BC + AB > 2CD

Hence proved!!! This proof is true for other medians in the triangle too.

Thank you...