Problem
1960 AHSME Problem 32
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:
![[asy] size(150); defaultpen(linewidth(0.8)+fontsize(10)); real e=350,c=55; pair O=origin,E=dir(e),C=dir(c),B=dir(180+c),D=dir(180+e), rot=rotate(90,B)*O,A=extension(E,D,B,rot); path tangent=A--B; pair P=waypoint(tangent,abs(A-D)/abs(A-B)); draw(unitcircle^^C--B--A--E); dot(A^^B^^C^^D^^E^^P,linewidth(2)); label("$O$",O,dir(290)); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,NE); label("$D$",D,dir(120)); label("$E$",E,SE); label("$P$",P,SW);[/asy]](/static/aops/31001830291c125521d5551ec5ec3990ff4aa5c9.png)
\begin{array}{l} \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} \end{array}
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