To verify that the sum of any two sides of a triangle is always greater
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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.
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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