Triangle Inequality Theorem

Consider a general triangle ABC and the three inequalities given by the theorem.

Here, it will be shown that AC+ AB > BC. The other two inequalities can be proved following the same procedure. Start by extending AB to the left of A. Then, consider a point D on this line such that AD= AC.

In △ ADC, the sides AD and AC are congruent. This means that by the Isosceles Triangle Theorem, the angles opposite them are congruent angles. Therefore, ∠ D ≅ ∠ DCA, which means that m∠ D = ∠ DCA.

Notice that ∠ DCB is made of ∠ DCA and ∠ ACB. Therefore, the measures of these three angles can be related thanks to the Angle Addition Postulate. m∠ DCB = m∠ DCA + m∠ ACB From this equation, m∠ DCB is greater than m∠ DCA. m∠ DCB &> m∠ DCA &⇓ m∠ DCB &> m∠ D Now consider △ DCB. From the previous inequality and the Triangle Larger Angle Theorem, the side opposite ∠ DCB is longer than the side opposite ∠ D. m∠ DCB &> m∠ D &⇓ DB &> BC The length of DB is equal to the sum of the lengths of DA and AB. This is because of the Segment Addition Postulate. Substituting the corresponding sum into the left-hand side results in the following inequality. DB &> BC &⇓ DA + AB &> BC Recall that D was plotted so that AD= AC. Therefore, substitute AC for DA into the last inequality. DA + AB &> BC &⇓ AC + AB &> BC ✓ Notice that the last inequality is the desired one. Consequently, the proof is complete.