Question

Prove that sum of any two sides of a triangle is greater than the third side.

Answer 1

Construction: In $\Delta ABC$, extend $AB$ to D in such a way that $AD=AC$.

In $\Delta DBC$, as the angles opposite to equal sides are always equal, so,

$\angle ADC=\angle ACD$

Therefore,

$\angle BCD>\angle BDC$

As the sides opposite to the greater angle is longer, so,

$BD>BC$

$AB+AD>BC$

Since $AD=AC$, then,

$AB+AC>BC$

Hence, sum of two sides of a triangle is always greater than the third side.