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

#1**+1 **

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

CPhill
Feb 28, 2018