Question

The sides of a quadrilateral are produced in order. What is the sum of the four exterior angles?

Answer 1

Let $ABCD$ be a quadrilateral such that exterior angles formed by extending the sides $AD,AB,BC$ and $CD$ are angles $x,y,z$ and $w$.

Now, since, sides $AD,AB,BC$ and $CD$ are straight lines, therefore,

$\Rightarrow \angle BAD+x={180}^{\circ }\Rightarrow \angle BAD={180}^{\circ }-x$...(i) [Linear Pair].

Similarly,

$\Rightarrow \angle ABC={180}^{\circ }-y$ ...(ii)

$\Rightarrow \angle BCD={180}^{\circ }-z$ ...(iii)

$\Rightarrow \angle ADB={180}^{\circ }-w$ ...(iv)

Adding (i), (ii), (iii) and (iv), we get,

$\Rightarrow \angle BAD+\angle ABC+\angle BCD+\angle ADB=({180}^{\circ }-x)+({180}^{\circ }-y)+({180}^{\circ }-z)+({180}^{\circ }-w)$

$\Rightarrow {360}^{\circ }={720}^{\circ }-(x+y+z+w)$ [Sum of all the angles of a quadrilateral is ${360}^{\circ }$]

$\Rightarrow x+y+z+w={360}^{\circ }$

Thus, the sum of the four exterior angles is ${360}^{\circ }$.

OR

It is well established that regardless of number of exterior angles, the sum of all the exterior angles of a polygon is always ${360}^{\circ }$.