Question

prove that the sum of any sides of a triangle is greater than the third side .. don't dare to spam ..

Answer 1

$\mathfrak{\underline{\underline{\red{Construction:}}}}$

${:⇒}$ In ΔABC, extend AB to D in such a way that AD=AC.

${:⇒}$In ΔDBC, as the angles opposite to equal sides are always equal, so,

${↪}$$\tt{∠ADC=∠ACD}$

Therefore,

${↪}$$\tt{∠BCD>∠BDC}$

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

$\tt\pink{BD >BC}$

$\tt\purple{AB+AD>BC}$

$\tt\orange{.°. AD=AC, then,}$

$\tt\green{AB+AC>BC}$

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