Given: ∆ABC is isosceles triangle.

To prove: ∠CAD = ∠CBE

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,

∠CAB +∠CAD = 180°

And ∠CAB +∠CBE = 180°

Which implies that,

∠CAD = 180° - ∠CAB

And ∠CBE = 180° - ∠CAB

Hence we say that ∠CAD = ∠CBE

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