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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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