Geometry Question: finding lengths in a triangle In a triangle $ABC, AB=AC$, $BC=22\sqrt3$, $\cos A=-\frac{17}{225}$. $D$ is a point on $AC$ such that $AD<DC$ and $P$ is the point on the segment $BD$ such that $\angle APC = 90^{\circ}$. Given $\angle ABD=\angle BCP$, find $BD$.
Here's a diagram of what it's like I guess: 
 
With reference to the diagram, $\cos (x+y)=\frac{11}{15}$, after taking $\cos (180^{\circ}-A)$ and so on. I kind of did cosine rule for both triangles $ABD$ and $BCD$ to get expressions for $BD^2$ and tried to sub in $\cos(x+y)$ but that is quite messy and doesn't really get anything. I also tried Stewart's theorem for $BD^2$. Are there any similar triangles or auxiliary lines that I'm not seeing? Can someone give a hint on this? Thanks.
 A: 
Hint:
\begin{align} 
\triangle BCD&\cong \triangle CPD
,\\
\frac{|BD|}{|CD|}
&=
\frac{|CD|}{|PD|}
=\frac{|BC|}{|PC|}
\end{align}
\begin{align} 
\cos\tfrac\alpha2&=\frac{2\sqrt{26}}{15}
=\sin\beta
,\\
\cos\beta&=\frac{11}{15}
,\\
|AB|=|AC|&=\frac{|BC|}{2\cos\beta}
=15\sqrt3
,\\
|AM|=|MC|=|MP|&=\tfrac12|AC|=\tfrac{15\sqrt3}2
,\\
\triangle CDP:\quad
\frac{|CD|}{|PD|}
&=\frac{\sin\beta}{\sin(\beta-\phi)}
,\\
\triangle PMD:\quad
|PD|
&=
\frac{|PC|\sin(\beta-\phi)}{\sin(2\beta-\phi)}
,\\
\frac{|BD|}{|CD|}
&=\frac{|BC|}{|PC|}
\\
&=\frac{|BC|}{|AC|\cos(\beta-\phi)}
=\frac{22}{15\cos(\beta-\phi)}
.
\end{align}
From
\begin{align} 
\frac{\sin\beta}{\sin(\beta-\phi)}
&=\frac{22}{15\cos(\beta-\phi)}
,\\
\phi&
=\beta-\arctan(\tfrac{15\sin\beta}{22})
=\beta-\arctan(\tfrac{\sqrt{26}}{11})
\end{align}
\begin{align} 
\frac{|BD|}{|CD|}
&=\frac{22}{15\cos(\arctan(\tfrac{\sqrt{26}}{11}))}
=\frac{14\sqrt3}{15}
,\\
\triangle CDP:\quad
|PD|
&=
\frac{|AC|\cos(\beta-\phi)\sin(\beta-\phi)}{\sin(2\beta-\phi)}
.
\end{align}
After simplification, $|PD|=\tfrac{75}7$
and then
\begin{align} 
\frac{|BD|}{|CD|}
\cdot
\frac{|CD|}{|PD|}
&=\left(\frac{|BD|}{|CD|}\right)^2
=\frac{196}{75}
=\frac{|BD|}{|PD|}
,\\
|BD|&=\frac{196}{75}
\cdot\frac{75}7
=28
.
\end{align}
A: We indicate with $\alpha$ the angle $\angle BAC$, so $cos(\alpha)=-\frac{17}{225}$.
The angle $\angle ABC=\frac{\pi}{2}-\frac{\alpha}{2}$ and the angle $\angle PCA=\angle PBC= (\frac{\pi}{2}-\frac{\alpha}{2})-x$ but $APC$ is rectangle and so $\angle PAC=\frac{\pi}{2}- \angle PCA= \frac{\pi}{2}- (\frac{\pi}{2}-\frac{\alpha}{2})+x$
$= \frac{\alpha}{2}+x $
So
$\angle PAD= \frac{\alpha}{2}+x $
And 
$\angle PAB=\alpha-(\frac{\alpha}{2}+x)=
\frac{\alpha}{2}-x$
And
$\angle PDA=\angle BDA=\pi-\alpha-x$
$PD=\frac{sin(\frac{\alpha}{2}+x)}{sin(\pi-\alpha-x)}AP= \frac{sin(\frac{\alpha}{2}+x)}{sin(\alpha+x)}AP=$
$ \frac{sin(\frac{\alpha}{2}+x)}{sin(\alpha+x)}sin(\frac{\alpha}{2}+x)AC= \frac{sin^2(\frac{\alpha}{2}+x)}{sin(\alpha+x)}AC= $
$\frac{sin^2(\frac{\alpha}{2}+x)}{sin(\alpha+x)}sin(\frac{\pi}{2}-\frac{\alpha}{2})BC $
and
$BP=\frac{sin(\frac{\alpha}{2}-x)}{sin(x)}AP= \frac{sin(\frac{\alpha}{2}-x)}{sin(x)} sin(\frac{\alpha}{2}+x)AC= $
$\frac{sin(\frac{\alpha}{2}-x)}{sin(x)} sin(\frac{\alpha}{2}+x)sin(\frac{\pi}{2}-\frac{\alpha}{2})BC $
