The exterior angle of a triangle

In seventh grade you learned that the angles of a triangle add up to $\displaystyle 180^{\circ}$. Angles between parallel lines show why.
We draw through the vertex $\displaystyle A$ a line $\displaystyle DE$ parallel to the side $\displaystyle BC$:

• $\displaystyle {\angle DAB}={\beta}$: they are alternate angles between the parallel lines $\displaystyle DE$ and $\displaystyle BC$.
• $\displaystyle {\angle EAC}={\gamma}$: alternate angles too.
• The three angles at $\displaystyle A$ – $\displaystyle \beta$, $\displaystyle \alpha$ and $\displaystyle \gamma$ – together make the straight line $\displaystyle DE$, so they add up to $\displaystyle 180^{\circ}$.

These are exactly the three angles of the triangle: $\displaystyle {\alpha+\beta+\gamma}={180^{\circ}}$.

We extend the side $\displaystyle BC$ beyond the vertex $\displaystyle C$, to the point $\displaystyle D$.

• $\displaystyle \delta$ is an exterior angle of the triangle: the angle between the side $\displaystyle AC$ and the extension of the side $\displaystyle BC$.
• The triangle’s angle $\displaystyle \gamma$ is adjacent to it, so $\displaystyle {\gamma+\delta}={180^{\circ}}$.
• The triangle’s other two angles, $\displaystyle \alpha$ and $\displaystyle \beta$, are the interior angles not adjacent to $\displaystyle \delta$.

Each vertex of the triangle has an exterior angle.

An exterior angle of a triangle equals the sum of the two interior angles not adjacent to it. In the figure above:

Both $\displaystyle \delta$ and $\displaystyle \alpha+\beta$ make up $\displaystyle 180^{\circ}$ with $\displaystyle \gamma$:

• $\displaystyle {\alpha+\beta+\gamma}={180^{\circ}}$, the sum of the angles of a triangle;
• $\displaystyle {\delta+\gamma}={180^{\circ}}$, since $\displaystyle \delta$ and $\displaystyle \gamma$ are adjacent.

So $\displaystyle {\delta}={180^{\circ}-\gamma}={\alpha+\beta}$.

In triangle $\displaystyle ABC$, $\displaystyle {\angle A}={50^{\circ}}$ and $\displaystyle {\angle B}={65^{\circ}}$.
We find the exterior angle $\displaystyle \angle ACD$ and the angle $\displaystyle \angle ACB$.

• $\displaystyle {\angle ACD}={50^{\circ}+65^{\circ}}={115^{\circ}}$, by the exterior angle theorem.
• $\displaystyle {\angle ACB}={180^{\circ}-115^{\circ}}={65^{\circ}}$, since it is adjacent to $\displaystyle \angle ACD$.

In triangle $\displaystyle ABC$, the side $\displaystyle BC$ is extended beyond $\displaystyle C$ to the point $\displaystyle D$, as in the figure above.
Find the angle asked for, and give the reasons.

$\displaystyle {\angle A}={40^{\circ}}$ and $\displaystyle {\angle B}={75^{\circ}}$. Find $\displaystyle \angle ACD$.

$\displaystyle {\angle ACD}={130^{\circ}}$ and $\displaystyle {\angle A}={55^{\circ}}$. Find $\displaystyle \angle B$.

$\displaystyle {\angle ACD}={100^{\circ}}$ and $\displaystyle {\angle A}={\angle B}$. Find $\displaystyle \angle A$.

In triangle $\displaystyle ABC$, the exterior angle at the vertex $\displaystyle C$ is $\displaystyle 3x+5$ degrees.
$\displaystyle {\angle A}={x+15}$ degrees, and $\displaystyle {\angle B}={50^{\circ}}$.
Write an equation and solve it.
Find the exterior angle and $\displaystyle \angle A$.

The figure shows two exterior angles of triangle $\displaystyle ABC$: $\displaystyle {\angle ACD}={135^{\circ}}$ and $\displaystyle {\angle EBA}={120^{\circ}}$.
Find $\displaystyle \angle A$, and give the reasons.

The figure shows one exterior angle at each vertex of triangle $\displaystyle ABC$: $\displaystyle \delta$, $\displaystyle \varepsilon$ and $\displaystyle \varphi$.
Explain why $\displaystyle {\delta+\varepsilon+\varphi}={360^{\circ}}$ in every triangle.
Hint: each exterior angle and the triangle’s angle adjacent to it make $\displaystyle 180^{\circ}$.