External Angle Of A Triangle

What is an external angle of a triangle? In the diagram below, in $\triangle ABC$ the side $BC$ is extended to the point $D$.

The angle formed by the extended line with the other line at point $C$, which is $AC$ is called the external angle. In this case the external angle is $\angle ACD$.

One of the very important properties of external angles is:

$\underline{The\ external\ angle\ is\ equal\ to\ the\ sum\ of\ the\ two\ opposite\ internal\ angles.}$

In the above diagram that would mean $\angle ACD = \angle ABC + \angle BAC$. We will see the formal proof of this now.

Given:

Any $\triangle ABC$ where side $BC$ has been extended to point $D$.

Required To Prove:

$\angle ACD = \angle ABC + \angle BAC$

Construction:

None.

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Proof:

$\angle BCD = 180^\circ$ because it is a straight line.

$\angle ACD + \angle ACB = \angle BCD = 180^\circ$

In $\triangle ABC$ sum of all the angles is $180^\circ$

$\texttip{\therefore}{therefore} \angle ABC + \angle BAC + \angle ACB = 180^\circ$

$\texttip{\therefore}{therefore} \angle ABC + \angle BAC + \angle ACB = \angle ACD + \angle ACB$

Subtracting $\angle ACB$ from both sides we get:

$\angle ABC + \angle BAC = \angle ACD$