Proof of the Exterior Angle Theorem.The exterior angle theorem for triangles states that the sum...

Step 1
In the $\mathrm{△}ABC$, we know that ABD is a straight line. So $\mathrm{\angle }ABC={180}^{\circ }-\mathrm{\angle }CBD$. From the angle sum property of triangles we can infer that $\mathrm{\angle }BAC+\mathrm{\angle }ABC+\mathrm{\angle }BCA={180}^{\circ }$ or $\mathrm{\angle }ABC={180}^{\circ }-(\mathrm{\angle }BAC+\mathrm{\angle }BCA)$.
Step 2
Therefore: $\mathrm{\angle }ABC={180}^{\circ }-\mathrm{\angle }CBD={180}^{\circ }-(\mathrm{\angle }BAC+\mathrm{\angle }BCA)$ $\Rightarrow -\mathrm{\angle }CBD=-(\mathrm{\angle }BAC+\mathrm{\angle }BCA)$ $\Rightarrow -\mathrm{\angle }CBD\times -1=-(\mathrm{\angle }BAC+\mathrm{\angle }BCA)\times -1$ $\Rightarrow \mathrm{\angle }CBD=\mathrm{\angle }BAC+\mathrm{\angle }BCA$

Step 1
Euclid gives a visually satisfying proof of the exterior angle theorem by drawing BE parallel to AC, and observing that $\mathrm{\angle }CBE=\mathrm{\angle }ACB$ (alternate interior angles) and $\mathrm{\angle }EBD=\mathrm{\angle }CAB$ (corresponding angles), making $\mathrm{\angle }CBD=\mathrm{\angle }ACB+\mathrm{\angle }CAB$. 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.

Step 2
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 $\mathrm{\angle }CBD=\mathrm{\angle }BCA+\mathrm{\angle }BAC$ and $\mathrm{\angle }ACE=\mathrm{\angle }CBA+\mathrm{\angle }BAC$ and $\mathrm{\angle }BAF=\mathrm{\angle }CBA+\mathrm{\angle }BCA$ then by addition $\mathrm{\angle }CBD+\mathrm{\angle }ACE+\mathrm{\angle }BAF=2\mathrm{\angle }BCA+2\mathrm{\angle }CBA+2\mathrm{\angle }BAC=2\cdot {180}^o={360}^o$.