Consider the following triangle, ∆ABC:
We need to prove that AB + AC > BC.
Proof:
Extend BA to point D such that AD = AC, and join C to D, as shown below:
We note that <ACD = <D, which means that in ∆ BCD, <BCD > <D. Sides opposite larger angles are larger, and thus: BD > BC
AB + AD > BC
AB + AC > BC (because AD = AC)
This completes our proof. We can additionally conclude that in a triangle:
Since the sum of any two sides is greater than the third, then the difference of any two sides will be less than the third.
The sum of any two sides must be greater than the third side.
