If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.


Given:

The base of an isosceles triangle is produced on both sides.

To do:

We have to prove that the exterior angles so formed are equal to each other.

Solution:

Let in an isosceles triangle $ABC, AB = AC$ and base $BC$ is produced both ways.

In $\triangle ABC$,

$AB = AC$

This implies,

$\angle ACB = \angle ABC$              (Angles opposite to equal sides are equal)

$\angle ACD + \angle ACB = 180^o$                    (Linear pair)

$\angle ABE + \angle ABC = 180^o$

Therefore,

$\angle ACD + \angle ACB = \angle ABE + \angle ABC$

$\angle ACD + \angle ACB = \angle ABE + \angle ACB$

This implies,

$\angle ACD = \angle ABE$

Hence proved.

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Updated on: 10-Oct-2022

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