Question

Prove that the sum of any two sides of a triangle is greater than the sum of its median formed on third side

Answer 1

Answer:

Given: A ∆ABC in which AD is median.

To Prove:

AC+AB>2AD

Construction:

Produce AD to E, such that AD=DE.

Join EC.

Proof:

In ∆ADB & ∆ EDC

AD=ED (by construction)

angleADB=angleEDC. (vertically opposite angle)

BD=CD. (D midpoint of BC)

∆ADB congruent ∆ EDC (by SAS)

AB=EC (by CPCT)

Now ,in ∆AEC, we have AC+EC>AE

[SINCE, SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER THAN THE THIRD SIDE]

AC+EC>AD+DE. (AE=AD+DE)

AC+AB>2AD (AD=ED & EC=AB)

Thus, the sum of any two sides of a triangle is greater than twice the median with respect to the third side.