Question

prove that the sum of any 2 side of a triangle is greater than the 3rd side

Answer 1

draw a triangle Abc
Given ΔABC, extend BA to D such that AD = AC.
Now in ΔDBC
∠ADC = ∠ACD [Angles opposite to equal sides are equal]
Hence ∠BCD > ∠ BDC
That is BD > BC [The side opposite to the larger (greater) angle is longer]
Þ AB + AD > BC 
∴ AB + AC > BC [Since AD = AC)
Thus sum of two sides of a triangle is always greater than third

Answer 2

Answer of dis question