Question

prove that in a triangle sum of any two 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.

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Answer 2

yes that is correct answer