Question

Prove that the sum of the angles of a triangle is ${180}^o$ and using this theorem prove that exterior angle is equal to the sum of interior opposite angle.

Answer 1

$△ABC$ is any triangle

$\angle BAC=\alpha ,\angle ABC=\beta ,\angle ACB=\gamma$

CA is extended and a line parallel

to BC and passing through A is

drawn as shown in figure

$\therefore$ From figure, it is obvious that $\alpha +\beta +\gamma ={180}^{\circ }$

$\therefore$ The sum of the angles of a triangle is always $={180}^{\circ }$

From the figure, it is also clear that $\beta +\gamma ={180}^{\circ }-\alpha$

where ${180}^{\circ }-\alpha$ is the exterior angle of the triangle