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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.
Given:
A $\triangle ABC$ in which AB, BC and CA are produced to points D, E and F.
To do:
We have to prove that the sum of the exterior angles so formed is equal to four right angles.
Solution:
We know that the measure of each exterior angle of a triangle is equal to the sum of the opposite and non-adjacent interior angles.
Therefore,
$\angle DCA=\angle A+\angle B$....(i)
$\angle FAE=\angle B+\angle C$.....(ii)
$\angle FBD=\angle A+\angle C$....(iii)
Adding equations (i), (ii) and (iii), we get
$\angle DCA+\angle FAE+\angle FBD=\angle A+\angle B+\angle B+\angle C+\angle A+\angle C$
$\angle DCA+\angle FAE+\angle FBD=2\angle A+2\angle B+2\angle C$
$\angle DCA+\angle FAE+\angle FBD=2(\angle A+\angle B+\angle C)$
The sum of all the angles in a triangle is $180^o$.
This implies,
$\angle DCA+\angle FAE+\angle FBD=2(180^o)$
$\angle DCA+\angle FAE+\angle FBD=4(90^o)$
Hence proved.
Updated on: 10-Oct-2022
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