Question

Prove that any two sides of a triangles are together greater than twice the median drawn to the third side .​

Answer 1

Answer:

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:

Pata nahi apki id ban kesay hojaty hai aur phir sahi bi hojaty hai aur itny jaldi itnay zyda points bi.

will u tell me ye kesay krlety ho?