Interior Angles Theorem

Consider a triangle with vertices $A,$ $B,$ and $C,$ and the parallel line to $\overline{BC}$ through $A.$ Let $\angle 1$ and $\angle 2$ be the angles outside $△ABC$ formed by this line and the sides $\overline{AB}$ and $\overline{AC}.$

By the Alternate Interior Angles Theorem, $\angle B$ is congruent to $\angle 1$ and $\angle C$ is congruent to $\angle 2.$

By the definition of congruent angles, $\angle 1$ and $\angle B$ have the same measure. For the same reason, $\angle 2$ and $\angle C$ also have the same measure. $$\begin{array}{cc}\angle B\cong \angle 1& \angle C\cong \angle 2\\ \Updownarrow & \Updownarrow \\ m\angle B=m\angle 1& m\angle C=m\angle 2\end{array}$$ Furthermore, in the diagram it can be seen that $\angle BAC,$ $\angle 1,$ and $\angle 2$ form a straight angle. Therefore, by the Angle Addition Postulate their measures add to $180^{\circ }.$ $$\begin{array}{c}m\angle BAC+m\angle 1+m\angle 2=180^{\circ }\end{array}$$ By the Substitution Property of Equality, it follows that the sum of the measures of $\angle BAC,$ $\angle B,$ and $\angle C$ is equal to $180^{\circ }.$ $$\begin{array}{c}m\angle BAC+m\angle 1+m\angle 2=180^{\circ }\\ \downarrow \\ m\angle BAC+m\angle B+m\angle C=180^{\circ }\end{array}$$ Finally, in $△ABC,$ $\angle BAC$ can be named $\angle A.$