Question

Prove that The sum of any two sides of a triangle is always greater than the third side​

Answer 1

Answer:

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.