Question

prove that the sum of any two sides of a triangle is greater than the third side ​

Answer 1

Answer , see to the diagram provided.

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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