Question

Prove that sum of angles of a triangle is $180°$

Answer 1

Let's take a triangle $△ABC$

Consider base $BC$, draw a line $MN$, parallel to $BC$

Now, $ABandAC$ will be transversal

$AB$ is transversal and $MN\parallel BC$

$\Rightarrow \angle 2=\angle 4$ (alternate interior angle)--------(1)

$AC$ is transversal and $MN\parallel BC$

$\Rightarrow \angle 3=\angle 5$(alternate interior angle)----------(2)

adding equations (1) and (2), we get

$\angle 2+\angle 3=\angle 4+\angle 5$

adding $\angle 1$ on both sides

$\angle 1+\angle 2+\angle 3=\angle 1+\angle 4+\angle 5$-------(3)

since,$\angle 1+\angle 4+\angle 5$ will form a linear pair.

As we know that the sum of angles of a linear pair is $180°$

$\Rightarrow \angle 1+\angle 4+\angle 5=180°$

substitute the value in equation(3). we get,

$$\begin{gathered} \Rightarrow \angle 1+\angle 2+\angle 3=180° \\ i.e.in△ABC,\angle A+\angle B+\angle C=180° \end{gathered}$$

Hence, proved.