Question

prove that in a triangle sum of 2 sides is greater than the third side

Answer 1

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 side.

Answer 2

Let the sides of triangle = 4.5 cm, 8cm, 6cm

nowlets check the property

4.5 + 8 > 6 
12.5 > 6 ----- (True) 

8 + 6 > 4.5 
14 > 4.5 ----- (True) 

6 + 4.5 > 8 
10.5 > 8 ----- (True) 

Hence, it's proved that "Sum of Two Sides of the Triangle is Always Greater than the Third Side.