1957 AHSME Problems/Problem 18

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Problem 18

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:

$[asy] defaultpen(linewidth(.8pt)); unitsize(2cm); pair O = origin; pair A = (-1,0); pair B = (1,0); pair C = (0,1); pair D = (0,-1); pair M = dir(45); pair P = intersectionpoint(O--C,A--M); draw(Circle(O,1)); draw(A--B); draw(C--D); draw(A--M); label("A",A,W); label("B",B,E); label("C",C,N); label("D",D,S); label("M",M,NE); label("O",O,NE); label("P",P,NW);[/asy]$

$\textbf{(A)}\ AO\cdot OB \qquad \textbf{(B)}\ AO\cdot AB\qquad \\ \textbf{(C)}\ CP\cdot CD \qquad \textbf{(D)}\ CP\cdot PD\qquad \textbf{(E)}\ CO\cdot OP$

Solution

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)}