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.
+130044 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
