Question

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

Answer 1

$$\begin{gathered} \mathrm{By}\mathrm{exterior}\mathrm{angle}\mathrm{property},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{TPR}=\angle \mathrm{PRQ}+\angle \mathrm{PQR}.....\left(\mathrm{i}\right) \\ \angle \mathrm{QRS}=\angle \mathrm{RPQ}+\angle \mathrm{PQR}......\left(\mathrm{ii}\right) \\ \angle \mathrm{UQT}=\angle \mathrm{RPQ}+\angle \mathrm{PRQ}.....\left(\mathrm{iii}\right) \\ \mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{exterior}\mathrm{angles}=\angle \mathrm{TPR}+\angle \mathrm{QRS}+\angle \mathrm{UQT} \\ \mathrm{From}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{get}: \\ \angle \mathrm{TPR}+\angle \mathrm{QRS}+\angle \mathrm{UQT}=\angle \mathrm{PRQ}+\angle \mathrm{PQR}+\angle \mathrm{RPQ}+\angle \mathrm{PQR}+\angle \mathrm{RPQ}+\angle \mathrm{PRQ} \\ \Rightarrow \angle \mathrm{TPR}+\angle \mathrm{QRS}+\angle \mathrm{UQT}=2(\angle \mathrm{PRQ}+\angle \mathrm{PQR}+\angle \mathrm{RPQ}) \\ \Rightarrow \angle \mathrm{TPR}+\angle \mathrm{QRS}+\angle \mathrm{UQT}=2\times 180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}\mathrm{of}△\mathrm{PRQ}) \\ \Rightarrow \angle \mathrm{TPR}+\angle \mathrm{QRS}+\angle \mathrm{UQT}=360°=4\mathrm{right}\mathrm{angles} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$