Question

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

Answer 1

Hint: Consider a triangle $ABC$. Extend $AB$ to $D$ such that $AD = AC$.Then apply the theorem that in a triangle , the side opposite to the larger angle is longer.

Complete step-by-step answer:
Given: A triangle $ABC$.

To prove: $AB + AC > BC$
$AB + BC > AC$
$BC + AC > AB$
Construction: Extend $AB$ to $D$ such that $AD = AC$.

Proof: In $\Delta ACD$,
$AC = CD$ (By Construction)
$\therefore \angle ADC = \angle ACD$ (In a triangle, angles opposite to equal sides are equal)
Now, $\angle BCD = \angle BCA + \angle ACD$
$\angle BCD = \angle BCA + \angle ADC$ $\left( {\because \angle ADC = \angle ACD} \right)$
$ \Rightarrow $$\angle BCD > \angle BDC$
$ \Rightarrow BD > BC$ (In a triangle , side opposite to the larger angle is longer)
$ \Rightarrow AB + AD > BC$ $\left( {\because BD = AB + AD} \right)$
$ \Rightarrow AB + AC > BC$ ($\because AD = AC$ by construction)

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

Note: The two other statements i.e., $AB + BC > AC$ and $BC + AC > AB$ can also be proved in the same manner as we proved $AB + AC > BC$.