Converse of the exterior angle bisector theorem How to prove the following theorem?
In triangle ABC the point P divides the extension of the line AB in the following ratio
AP:BP=AC:BC.
Prove that the line CP is the bisector of the exterior angle C.
Trigonometric solution is possible.
 A: 
$\mathrm{Fig. 1}$ shows the triangle $ABC$ and the line $CP$ mentioned in your question. 
We need to draw the line $BD$ parallel to $PC$ to facilitate our proof. We can write right away,
$$\frac{AP}{BP}=\frac{AC}{CD}. \tag{because PC $\backslash\backslash$ BD}$$
But, it is given that,
$$\frac{AP}{BP}=\frac{AC}{BC}.$$
Therefore,
$$\frac{AC}{CD}=\frac{AC}{BC},$$
which means $CD=BC$. That makes $BCD$ an isosceles triangle and , as a consequence,
$$\measuredangle DBC=\measuredangle CDB.\tag{1}$$
Because they are alternate angles, The two angles $\measuredangle BCP$ and $\measuredangle DBC$ are equal, i.e.
$$\measuredangle PCB=\measuredangle DBC.$$
The two angles $\measuredangle ECP$ and $\measuredangle CDB$ are corresponding angles. That makes them equal angles too, i.e.
$$\measuredangle ECP =\measuredangle CDB.$$
Finally, because of the identicalness seen in the equation(1), we have, 
$$\measuredangle PCB =\measuredangle ECP.$$
$\underline{Note}$:
You could have refer to a good book on elementary geometry for this proof, because they usually provide the proofs of the
 exterior angle bisector theorem and its converse side by side under the heading $Euclid\space VI\space 3$. If you have done that, you could have avoided this days-long waiting.
