Circle $O$ has diameters $AB$ and $CD$ perpendicular to each other. $AM$ is any chord intersecting $CD$ at $P$. 
Then $AP\cdot AM$ is equal to: 

$\textbf{(A)}\ AO\cdot OB \qquad \textbf{(B)}\ AO\cdot AB\qquad \\ \textbf{(C)}\ CP\cdot CD \qquad  \textbf{(D)}\ CP\cdot PD\qquad  \textbf{(E)}\ CO\cdot OP$
Draw $MB$. Since $\angle AMB$ is inscribed on a diameter, $\angle AMB$ is $90^\circ$. By AA Similarity, $\triangle APO ~ \triangle ABM$. Setting up ratios, we get $\frac{AP}{AO}=\frac{AB}{AM}$. Cross-multiplying, we get $AP\cdot AM = AO \cdot AB$, so the answer is $\textbf{(B)}$