First, consider a segment $\overline{AD}$ from $A$ to the side $\overline{BC}$ 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 $\angle BAD$ and $\angle BDA$ are congruent angles.

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