Theorem 3

Theorem 3

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

Given: $\triangle ABC$

To prove: $\text{i. }m\angle \overline{AB}+\overline{AC}$> $m \overline{BC}$
$\text{ii. }m\angle \overline{AB}+\overline{BC}$> $m \overline{CA}$ $\text{iii. }m\angle \overline{AC}+\overline{BC}$>$m \overline{AB}$

Construction: Produce $\overrightarrow{BA}$ to $D$, making $\overline AD \cong \overline{AC}$ Draw $\overline{DC}$.

Explanation

• To prove that the sum of the lengths of any two sides of a triangle is greater than the length of the third side, we start by considering triangle $\triangle ABC$. First we will extend side $\overline{AB}$ to $D$ such that $\overline{AD}\cong\overline{AC}$, which implies that $\angle1$ and $\angle2$ are congruent, since they are opposite to these sides.

• We also have $\angle BCD = \angle BCA + \angle 1$. Now, since $\angle BCD$ is an exterior angle of triangle $\triangle ABC$, it is greater than the non-adjacent interior angle $\angle 1$. Therefore, we have $\angle BCD$ > $\angle 2$, by transitivity of inequality.

• This implies that in triangle $\triangle BDC$, the side opposite to the greater angle $\angle BCD$ is longer. Thus, we have $m\overline{BD}$ > $m\overline{BC}$. However, we know that $\overline{BD}=\overline{AB}+\overline{AD}$. Substituting the value of $\overline{AD}$ from the construction, we get $\overline{BD}=\overline{AB}+\overline{AC}$.

• Therefore, we have $m\overline{AB}+m\overline{AC}$ > $m\overline{BC}$, since $m\overline{BD}$>$m\overline{BC}$. Similarly, we can show that $m\overline{AB}+m\overline{BC}$>$m\overline{AC}$ and $m\overline{BC}+m\overline{AC}$>$m\overline{AB}$ using the same process.

• Hence, we have shown that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.