Find the $\angle ACB$ of $\triangle ABC$. 
If $PC=2BP$, $\angle ABC= 45^\circ$, and $\angle APC=60^\circ$, find $\angle ACB$. 


All solutions are acceptable but please try solving using reflection of point $C$ through the line segment $AP$.

I found a solution by adding a dot named D but in the given solution which I've drawn, I didn't understand how alpha was found. Although it is the original given solution in the geometry book. 

It was only pointed that 
$\angle ACB = \angle ADP = \frac{1}{2}(180^\circ - \angle BDP) = 75^\circ$.
Please explain this for me because this solution is blurry and vague for me.
 A: 
Sometimes, geometry consists of dropping the right line or introducing the accurate point...


Denote by $D$ the point on $AP$ such that $\angle PDC=90°$ and let $BP=x$. Note now that the triangle $\Delta PCD$ is a $30°-60°-90°$ triangle, which implies that $PD=x$. 
Therefore $\Delta BPD$ is isosceles and hence $\angle PBD=30° \Rightarrow \angle DBA=15°$.
Easy angle chasing leads to the conclusion that $DC=DB=DA$, which implies that $\Delta CDA$ is also isosceles. Thus $\angle ACD=45°$. $$\therefore \angle ACB=45°+30°=75° \blacksquare$$
A: By calculating some angles and applying the sine rule we have:
$$|AB|=\frac{3\sqrt{2}+\sqrt{6}}{2}\cdot x$$
Then a second application of the sine rule on $\triangle ABC$ gives:
$$\frac{3x}{\sin{(135^\circ-\angle ACB)}}=\frac{\frac{3\sqrt{2}+\sqrt{6}}{2}\cdot x}{\sin{(\angle ACB)}}$$
$$\frac{3}{\sin{(135^\circ-\angle ACB)}}=\frac{3\sqrt{2}+\sqrt{6}}{2\sin{(\angle ACB)}}$$
The only solution in $0^\circ \lt\angle ACB \lt 180^\circ$ to the above equation is then:
$$\angle ACB =75^\circ$$
A: Hint: Drop an altitude from $A$ to $\overline{BC}$ and then use trigonometry.
