Question

The side BC of $△ABC$ is produced on both sides. Prove that the sum of the exterior angles so formed is greater than $\angle A$ by two right angles.

Answer 1

In $\Delta$ABC

$\angle A+\angle B+\angle C={180}^o$ ………$\left(1\right)$

Now,

$\angle B+\angle ABP={180}^o$

(linear pair)

$\angle C+\angle ACQ={180}^o$ (linear pair)

$\Rightarrow \angle B+\angle C+\angle ABP+\angle ACQ={360}^o$

$\angle B+\angle C=180-\angle A$ (from $\left(1\right)$)

$\Rightarrow 180-\angle A+\angle ABP+\angle ACQ={360}^o$

$\Rightarrow \angle ABP+\angle ACQ={360}^o-{180}^o+\angle A$

$=180+\angle A$

$\Rightarrow$ Sum of exterior angles is ${180}^o$ ($2$ right angles) more than $\angle A$.