Question

Prove that the sum of all interior angles of a triangle is 180°.

Answer 1

Let ABC be a triangle. Draw a line parallel to AC and passing hrough B as shown in the figure.

Since $AC\left|\right|DE$, therefore
$$\begin{gathered} \angle DBA=\angle BAC.....\left(i\right)\left(\mathrm{Alternative}\mathrm{angles}\right) \\ \angle EBC=\angle ACB.....\left(ii\right)\left(\mathrm{Alternative}\mathrm{angles}\right) \end{gathered}$$
Here, DBE is a straight line, so
$$\begin{gathered} \angle DBA+\angle ABC+\angle EBC={180}^{\circ } \\ \angle BAC+\angle ABC+\angle ACB={180}^{\circ }\left[\mathrm{From}\left(i\right)\mathrm{and}\left(ii\right)\right] \end{gathered}$$
Hence proved.