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Let B, C, and D be points on a circle. Let line BC and the tangent to the circle at D intersect at A. If AB = 4, AD = 8, and line AC is perpendicular to line AD, then find CD.

 

Guest Feb 28, 2018
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We can use the Secant-Tangent Theorem first to find BC

 

Specifically

 

The length of the tangent squared  =  the product of length of the tangent * its exterior segment

 

8^2  =  (4 + BC) * 4  simplify

 

64  =  16 + 4BC         subtract  16 from both sides

 

48  =  4BC    divide both sides by 4

 

12  =  BC

 

So......AD  = 8  and  AC =   4 + BC  =  16

 

And these will form two legs of a right triangle with CD as a hypotenuse

 

So

 

CD   =  √  ( 8^2 + 16^2)  = √ [ 8^2 ( 1 + 2^2) ]  =   8√5   units

 

 

cool cool cool

CPhill  Feb 28, 2018

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