Triangle Longer Side Theorem

First, consider a segment AD from A to the side BC so that BD is equal to AB. With this, an isosceles triangle △ ABD is created.

By the Isosceles Triangle Theorem, it can be stated that ∠ BAD and ∠ BDA are congruent angles.

Notice that, by the Angle Addition Postulate, m∠ BAC can be written as the sum of m∠ BAD and m∠ DAC. m∠ BAC=m∠ BAD+m∠ DAC Since ∠ BAD is congruent to ∠ BDA, their measures are equal. Therefore, m∠ BDA can be substituted for m∠ BAD in this equation. Additionally, in △ ADC, note that ∠ BDA is the exterior angle to the non-adjacent angles ∠ DAC and ∠ C. Therefore, by the Triangle Exterior Angle Theorem, its measure is equal to the sum of the measures of ∠ DAC and ∠ C. m∠ BDA=m∠ DAC+m∠ C Next, m∠ DAC+m∠ C can be substituted for m∠ BDA into the previously obtained equation. Recall that the measure of every angle, including ∠ DAC, is a non-negative value. Furthermore, observing the diagram, it can be seen that the measure of ∠ DAC is not 0^(∘). Under these conditions, it is implied that the measure of ∠ BAC is greater than the measure of ∠ C. m∠ BAC=2m∠ DAC+m∠ C ⇓ m∠ BAC > m∠ C Therefore, ∠ BAC, which is opposite the longer side BC, is larger than ∠ C, which is opposite the shorter side AB. This concludes the proof.