Lemma: $ \{X \in (0, \pi) | f(X) = \frac{\sin (\Phi-X)}{\sin X} \}$ is decreasing for any $\Phi \in (0,\pi),$ because its derivative $f'(X) = -\frac{\sin \Phi}{\sin^2 X} < 0.$ $CA + \frac{_1}{^2} AB = BC \Longleftrightarrow$ $\frac{3}{4} \cdot \frac{AB}{CA} = \frac{CA^2 + CA \cdot AB + AB^2 - BC^2}{CA \cdot AB} \Longleftrightarrow$ $\frac{\sin \angle ACP}{\sin \angle CPA} = \frac{AP}{CA} = 1 + 2\cos A = 3 - 4 \sin^2 \frac{A}{2} = \frac{\sin \frac{3A}{2}}{\sin \frac{A}{2}} = \frac{\sin (\pi - \frac{3A}{2})}{\sin \frac{A}{2}}.$ In addition, $\angle ACP + \angle CPA = \pi - A = \pi - \frac{3A}{2} + \frac{A}{2}.$ By the lemma, $\angle CPA = \frac{A}{2}.$

Take $M\in(BC)$ such that $CA=CM$. Let $CA=CM=b$, $BP=\frac13AP=\frac12BM=a$. By Stewart's theorem, \[CP^2=CB^2\cdot\frac34+CA^2\cdot\frac14-AP\cdot PB=(b+2a)^2\cdot\frac34+b^2\cdot\frac14-3a^2=b(b+3a)=CA(CA+AP).\] It is well-known that $\angle CAP=2\angle CPA\Leftrightarrow CP^2=CA(CA+AP)$, hence our proof is complete.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=371578

it hsas the following nice analytical form: let C be a point on the left branch of a hyperbola with focuses A,B,P is its right vertex. prove that $\angle CAP=2\angle CPA$.

Johan Gunardi wrote: $\angle CAP=2\angle CPA\Leftrightarrow CP^2=CA(CA+AP)$ Where does this come from?

silvergrasshopper wrote: Johan Gunardi wrote: $\angle CAP=2\angle CPA\Leftrightarrow CP^2=CA(CA+AP)$ Where does this come from? see here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=54513&

Or : calculate PC using Stewart's Theorem, then apply sinus theorem in APC. The problem becomes equivalent with $ 2 \cos ( \frac{A}{2})=\frac{PC}{AC}$. Square this relation, and use one of Neper Formulas for cos(A/2) : $ \cos ( \frac{A}{2})= \sqrt{\frac{p(p-a)}{bc}}$.

You may see here as well. Best regards, sunken rock