Question

If the sides of a triangle are produced in order, prove that the sum of the exterior angles so formed is equal to four right angles.

Answer 1

Prove the required statement:

Let a triangle $ABC$ in which the sides are$AB,BC,CA$ produced $F,D$ and $E$.

We have to prove $\angle DCA+\angle FAE+\angle FBD=360°$.

By using the theorem, an exterior angle of a triangle is equal to the sum of two remote interior angles.

From the Figure, we can write,

$\angle A+\angle B=\angle DCA$ . . . . . . $\left(1\right)$

$\angle B+\angle C=\angle FAE$ . . . . . . $\left(2\right)$

$\angle A+\angle C=\angle FBD$ . . . . . . $\left(3\right)$

Adding all three equations, we get

$\angle A+\angle B+\angle B+\angle C+\angle A+\angle C=\angle DCA+\angle FAE+\angle FBD$

$2\angle A+2\angle B+2\angle C=\angle DCA+\angle FAE+\angle FBD$

$2(\angle A+\angle B+\angle C)=\angle DCA+\angle FAE+\angle FBD$ . . . . . . . $\left(4\right)$

Since we know that sum of a triangle is $180°$

So, $\angle A+\angle B+\angle C=180°$

By putting the value in equation $\left(4\right)$, we get

$2(180°)=\angle DCA+\angle FAE+\angle FBD$

$360°=\angle DCA+\angle FAE+\angle FBD$

$90°\times 4=\angle DCA+\angle FAE+\angle FBD$

$4rightangles=\angle DCA+\angle FAE+\angle FBD$

Hence, proved that the sum of the exterior angles so formed is equal to four right angles.