3.07 Exterior angles of a triangle

If we extend a side segment of a triangle beyond the vertex, a new angle is formed. This angle is called an exterior angle:

All highlighted angles are exterior angles

Using what we know about parallel lines will allow us to prove the following interesting and useful fact:

Let's prove this by considering a generic triangle $\Delta ABC$ΔABC:

We start by forming an exterior angle at $A$A by extending $\overline{BA}$BA to $\overrightarrow{BA}$›‹BA, and mark a point $P$P  on the ray for referring to angles later, highlighting the exterior angle:

We also extend $\overline{BC}$BC to $\overrightarrow{BC}$›‹BC, and draw a parallel ray emanating from $A$A (marking any reference point $Q$Q):

We now have two parallel rays, $\overrightarrow{BC}$›‹BC and $\overrightarrow{AQ}$›‹AQ. Considering $\overline{BA}$BA as a transversal to these rays, we can conclude by the corresponding angles theorem that $\angle CBA\cong\angle QAP$∠CBA≅∠QAP:

Similarly, considering $\overline{CA}$CA as a transversal, we conclude by the alternate interior angles theorem that $\angle ACB\cong\angle CAQ$∠ACB≅∠CAQ:

The angle addition theorem then tells us that, since $m\angle CAP$m∠CAP, the exterior angle, is equal to $m\angle CAQ+m\angle QAP$m∠CAQ+m∠QAP, we can conclude that $m\angle CAP=m\angle ACB+m\angle CBA$m∠CAP=m∠ACB+m∠CBA by substitution, QED.

We will use this result to find unknown angles within triangles.

Worked example

Consider the following triangle:

Find the measure of $\angle LKM$∠LKM.

Think: This problem involves an exterior angle, so we will use the theorem above.

Do: Write the equation $m\angle KLM+m\angle LKM=m\angle KMQ$m∠KLM+m∠LKM=m∠KMQ, and substitute in the given measures to produce $58^\circ+m\angle LKM=131^\circ$58°+m∠LKM=131°. After subtracting $58^\circ$58° from both sides we find $m\angle LKM=73^\circ$m∠LKM=73°.