Exercise 13 Page 394

We are given the following figure.

We want to write an informal proof to show that an exterior angle of a triangle is equal to the sum of its two remote interior angles. Let's do it! First, notice that $\angle 1$ and $\angle 4$ form a linear pair. As a result, their angle measures add up to $180^{\circ }.$ Now, see that by the Subtraction Property of Equality we can isolate $m\angle 1$ on one side of the equality. Moving on, by the Angle Sum of a Triangle Theorem the sum of the interior angles of any triangle is $180^{\circ }.$ This means that the measures of $\angle 2,$ $\angle 3,$ and $\angle 4$ add up to $180^{\circ }.$ This time we are going to isolate $m\angle 2+m\angle 3$ one one side. The equality is still true by the Subtraction Property of Equality. Last, recall that $m\angle 1$ equals $180^{\circ }-m\angle 4.$ This means we can substitute $m\angle 1$ for $180^{\circ }-m\angle 4$ in the equality. We found that the exterior angle of the triangle, $\angle 4,$ has the same measure as the sum of the remote interior angles, $\angle 2$ and $\angle 3.$ Let's now summarize our proof in one paragraph.

Since $\angle 1$ and $\angle 4$ form a linear pair, $m\angle 1+m\angle 4=180^{\circ }.$ By the Subtraction Property of Equality, $m\angle 1=180^{\circ }-m\angle 4.$ Since $ABC$ is a triangle, $m\angle 2+m\angle 3+m\angle 4=180^{\circ }.$ By the Subtraction Property of Equality, $m\angle 2+m\angle 3=180^{\circ }-m\angle 4.$ So, by substitution, $m\angle 2+m\angle 3=m\angle 1.$