Question

If the sides of a triangle are produced in order, prove that the sum of the exterior angles so formed is equal to four right angles.

Answer 1

Side BC of triangle ABC is produced to D.
$\angle ACD=\angle B+\angle A...\left(i\right)$
Side AC of triangle ABC is produced to E.
$\angle BAC=\angle B+\angle C...\left(i\right)$
And side AB of triangle ABC is produced to F.
$\angle CBF=\angle C+\angle A...\left(iii\right)$

$$\begin{gathered} \mathrm{Adding}\left(i\right),\left(ii\right)\mathrm{and}\left(iii\right),\text{we get:} \\ \angle ACD+\angle BAE+\angle CBF=2\left(\angle A+\angle B+\angle C\right) \end{gathered}$$
$$\begin{gathered} =2\left(180\right)° \\ =360°=4\times 90°=4\mathrm{right}\mathrm{angle}s \end{gathered}$$
Hence, the sum of the exterior angles so formed is equal to four right angles.