Finding a triangle ratio. 
In the triangle ABC, the point P is found on the side AB. AC = 6 cm, AP = 4 cm, AB = 9 cm. Calculate BC:CP.

For some reason, I cannot get this even though I tried for half an hour.
I got that, $BC/CP < 17/10 = 1.7$ by the triangle inequality. $AP/PB = 4/5$
But that does not help one but, I'm very stuck!

Notice, let $BC=x$ & $CP=y$ now, applying cosine rule in $\triangle ABC$ $$\cos\angle BAC=\frac{AB^2+AC^2-BC^2}{2(AB)(AC)}=\frac{9^2+6^2-y^2}{2(9)(6)}=\frac{117-y^2}{108}\tag 1$$
similarly, applying cosine rule in $\triangle APC$
$$\cos\angle BAC=\frac{AP^2+AC^2-PC^2}{2(AP)(AC)}=\frac{4^2+6^2-x^2}{2(4)(6)}=\frac{52-x^2}{48}\tag 2$$
Equating (1) & (2), we get 
$$\frac{117-y^2}{108}=\frac{52-x^2}{48}$$ 
$$468-4y^2=468-9x^2$$
$$\frac{y^2}{x^2}=\frac{9}{4}\implies \frac{y}{x}=\frac{3}{2}$$
$$\color{red}{BC:CP=3:2}$$
