Question

Prove that the sum of angles of a quadrilateral is ${360}^o$

Answer 1

Let ABCD be any quadrilateral divided into four triangles, then using sum of angles in a $\Delta$ is ${180}^o\Rightarrow$

$\angle 1+\angle 2+\angle 3={180}^o$

$\angle 4+\angle 5+\angle 6={180}^o$

$\angle 7+\angle 8+\angle 9={180}^o$

$\angle 10+\angle 11+\angle 12={180}^o$

$\therefore \angle 1+\angle 2+\angle 3+...\angle 12=180\times 4$ ...(1)

Now

$\angle 3+\angle 6+\angle 7+\angle 10={360}^o$

equation (1) - equation (2) $\Rightarrow$

$\angle 1+\angle 2+\angle 4+\angle 5+\angle 8+\angle 9+\angle 11+\angle 12=180\times 4-360$

$\Rightarrow$ Sum of angles in a quadrilateral = $180\times 4-360={360}^o$

Hence proved