Question

prove that sum of any two side of triangle is greater than twice the medium draw to third side​

Answer 1

Given: In △ABC,AD is the median drawn from A to BC
To Prove that AB+AC>AD
Construction:Produce AD to E so that DE=AD, join BE
Proof:In △ADC and △EDB we have
AD=DE(constant)
DC=BD as D is the midpoint
∠ADC=∠EDB (vertically opposite angles)
∴ In △ABE, △ADC≅△EDB by S.A.S
This gives BE=AC
AB+BE>AE
AB+AC>2AD ∵AD=DE and BE=AC
Hence the sum of any two sides of a triangle is greater than twice the median with respect to the third side.

Answer 2

Answer:

Given

ABC is a triangle. AD is the median drawn on side BC. {Since, BD=DC}

To Prove

AB+AC=2AD

Proof

Since,Sum of two sides of a triangle is greater than the third side.

so,

In triangle ABD

AB+BD>AD ––inequality 1.

In triangle ADC

AC+DC>AD––inequality 2.

Now, adding both inequality we get,

AB+BD+AC+DC >AD+AD (Since BD=DC)

AB+BD+AC+DC>2AD

AB+AC>2AD. (therefore,BD&DC gets cancelled)

hence proved that the sum of two sides of a ∆is greater than the median drawn to the third side

(AB+AC>2AD)