In this figure the center of the circle is $O$. $AB \perp BC$, $ADOE$ is a straight line, $AP = AD$, and $AB$ has a length twice the radius. Then:

$\textbf{(A)} AP^2 = PB \times AB\qquad \\ \textbf{(B)}\ AP \times DO = PB \times AD\qquad \\ \textbf{(C)}\ AB^2 = AD \times DE\qquad \\ \textbf{(D)}\ AB \times AD = OB \times AO\qquad \\ \textbf{(E)}\ \text{none of these}$
We claim that $\fbox{A}$ is the right answer.
Let the radius of circle $O$ be $r$, and let the length of $AD=x$. Since $AB \perp BC$, $AB$ is a tangent to circle $O$. Thus, by the tangent-secant theorem, we have $AB^2=AD\times AE$, or, $(2r)^2=x(x+2r)$. Through some algebraic manipulation, we find \begin{align}AD^2=x^2=2r(2r-x)\end{align}. 
Since $AD^2=AP^2=x^2$, $PB=AB-AP=2r-x$, and $AB=2r$, we see that $(1)$ is identical to $AP^2=PB\times AB$, hence our answer is $\fbox{A}$