Let $w$ be a circle, and let $P$ be a point outside $w$. Let $X, Y$ be the tangents from $P$ to $w$. A line from $P$ intersects $w$ in two points $B, D$. Let $C$ be the intesection of $\overline{XY}$ and $\overline{BC}$. Prove that $CD\cdot BP=BC \cdot DP$.
I've tried using power of the point repeatedly but just get a mess. Cross ratios can be computed for four collinear points (and the relation we are asked to prove is that the cross ratio (ignoring signs) is $1$) but I don't know how to use that.
Denote $PX=PY=t$, $CD=a, BC=b, PD=c, CX=x, CY=y$.
By the power of point $C$ we have \begin{equation} ab=xy. \end{equation}
By the power of point $P$ we have
\begin{equation} t^2=c(a+b+c). \end{equation}
Finally, apply Stewart's theorem in triangle $PXY$ for the cevian $PC$. We obtain:
$PC^2\cdot XY+CX\cdot CY \cdot XY=PX^2\cdot CY+PY^2\cdot CX$ from which
$(c+a)^2\cdot(x+y)+xy(x+y)=t^2(x+y)$ and after dividing by $x+y$ it follows that
\begin{equation} t^2=(c+a)^2+xy. \end{equation}
Combining the three equations above we obtain $a(a+b+c)=bc$ which is the desired equality.