Problem

1997 AHSME Problem 26

Triangle ABC and point P in the same plane are given. Point P is equidistant from A and B, angle APB is twice angle ACB, and \overline{AC} intersects \overline{BP} at point D. If PB = 3 and PD= 2, then AD\cdot CD =

[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3,1); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); dot(A);dot(B);dot(C);dot(P);dot(D); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE + N);label("$P$",P,N); draw(A--B--P--cycle); draw(A--C--B--cycle);[/asy]

\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9


Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.


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