Proof of the Exterior Angle Theorem The exterior angle theorem for triangles states that the sum of “The measure of an exterior angle of a triangle is equal to the sum of the measures of the two interior angles that are not adjacent to it; this is the exterior angle theorem. The sum of the measures of the three exterior angles (one for each vertex) of any triangle is $360$ degrees.” How can we prove this theorem?

So, for the triangle above, we need to prove why $\angle CBD = \angle BAC + \angle BCA$.
 A: In the triangle $\triangle ABC$, we know that $ABD$ is a straight line. So $\angle ABC = 180^{\circ} - \angle CBD$. From the angle sum property of triangles we can infer that $\angle BAC + \angle ABC + \angle BCA = 180^{\circ}$ or $\angle ABC = 180^{\circ}-(\angle BAC + \angle BCA)$. Therefore:
$$\angle ABC = 180^{\circ} - \angle CBD = 180^{\circ} - (\angle BAC + \angle BCA)$$
$$\Rightarrow - \angle CBD = -(\angle BAC + \angle BCA)$$
$$\Rightarrow - \angle CBD \times -1 = -(\angle BAC + \angle BCA) \times -1$$
$$\Rightarrow \angle CBD = \angle BAC + \angle BCA$$
A: In Elements I, 32 Euclid gives a visually satisfying proof of the exterior angle theorem by drawing $BE$ parallel to $AC$, and observing that $\angle CBE=\angle ACB$ (alternate interior angles) and $\angle EBD=\angle CAB$ (corresponding angles), making$$\angle CBD=\angle ACB+\angle CAB$$ This theorem includes the further important result that the three angles of a triangle sum to $180^o$, or "two right angles" as Euclid says.

But if, as I suspect, the true intent of OP's question is, assuming the truth of the exterior angle theorem, prove that the sum of the three exterior angles of a triangle is $360^o$, then we can argue as follows.

Since by the exterior angle theorem$$\angle CBD=\angle BCA+\angle BAC$$and$$\angle ACE=\angle CBA+\angle BAC$$and$$\angle BAF=\angle CBA+\angle BCA$$then by addition$$\angle CBD+\angle ACE+\angle BAF= 2\angle BCA+2\angle CBA+2\angle BAC=2\cdot 180^o=360^o$$
