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

1957 AMC 12 Problem 18

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18.

Circle

O

has diameters

\overline{AB}

and

\overline{CD}

perpendicular to each other.

\overline{AM}

is any chord intersecting

\overline{CD}

P.

Then

AP\cdot AM

is equal to:

AO\cdot OB

AO\cdot AB

CP\cdot CD

CP\cdot PD

CO\cdot OP

Because

AB

is a diameter, angle

AMB

is a right angle

Compare right triangles

APO

and

ABM

Angles

APO

and

ABM

are both right angles, and the two triangles share angle

A.

Therefore triangles

APO

and

ABM

are similar. Corresponding sides give

\frac{AP}{AB}=\frac{AO}{AM}.

Hence

AP\cdot AM=AO\cdot AB.

Thus, the correct answer is B.