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

Given: $△ABC$ in which $AB,BC$ and $CA$ are produced to points $F,D$ and $E$ respectively as shown in the following figure.

To prove: $\angle DCA+\angle FAE+\angle FBD=4\times {90}^{\circ }$
$={360}^{\circ }$
Proof :
From the figure, using Exterior Angle Property, we get,
$\angle DCA=\angle CAB+\angle ABC...\left(1\right)$

$\angle FAE=\angle ABC+\angle BCA...\left(2\right)$

$\angle FBD=\angle CAB+\angle BCA...\left(3\right)$

By adding equation $\left(1\right),\left(2\right)$ and $\left(3\right)$ we get

$\angle DCA+\angle FAE+\angle FBD=\angle CAB+\angle ABC+\angle ABC+\angle BCA+\angle CAB+\angle BCA$

So we get.
$\angle DCA+\angle FAE+\angle FBD=2\angle CAB+2\angle ABC+2\angle BCA$

Now by taking out $2$ as common

$\angle DCA+\angle FAE+\angle FBD=2(\angle CAB+\angle ABC+\angle BCA)$

We know that the sum of all the angles in a triangle is ${180}^{\circ }$.

So we get,

$\angle DCA+\angle FAE+\angle FBD=2\left({180}^{\circ }\right)$

$\angle DCA+\angle FAE+\angle FBD={360}^{\circ }=4\times {90}^{\circ }$

Therefore, it is proved.