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Given: ∆ABC is isosceles triangle.

Let ∆ABC be our isosceles triangle as shown in the figure. We know that base angles of the isosceles triangle are equal.

Here, CAB = CBA ….(1)

Also here, CAD and CBE are exterior angles of the triangle.

So, we know that,

CAB +CAD = 180°… exterior angle theorem

And CBA + CBE = 180° … exterior angle theorem

So from (1) and above statement, we conclude that,

And CAB +CBE = 180°

Which implies that,

And CBE = 180° - CAB

Hence we say that CAD = CBE

For the isosceles triangle, the exterior angles so formed are equal to each other.

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