Let B, C, and D be points on a circle. Let BC and the tangent to the circle at D intersect at A. If AB = 4, AD = 8, and AC is perperpendicular to AD, then find CD.
+118608 I do not have time to do it all right now but it is easy enough anyway.
use the rule below:
Let BC= x
set up a quadric equation.
Find x using the quadratic formula
Find CD using the ythagorean theorum.
All done 
The rule was copied from this site:
+128460 We have, using the secant-tangent theorem, that
AC * AB = AD^2
(BC + AB) * AB = AD^2
(BC + 4) * 4 = 8^2
4BC + 16 = 64 divide through by 4
BC + 4 = 16 subtract 4 from each side
BC = 12
So AC = (12 + 4) = 16
Using the Pythagorean Theorem to find CD, we have
AC * AB = AD^2
AC * 4 = 64
AC = 16
Therefore
AC^2 + AD^2 = CD^2
sqrt (AC^2 + AD^2) = CD
sqrt (16^2 + 8^2) = CD
sqrt ( 256 + 64) =
sqrt ( 64 [ 4 + 1 ] ) =
8 sqrt (5) ≈ 17.889
+26367 Let B, C, and D be points on a circle. Let BC and the tangent to the circle at D intersect at A.
If AB = 4,
AD = 8,
and AC is perperpendicular to AD,
then find CD.
\(\begin{array}{|rcll|} \hline AD^2 &=& AB \cdot AC & \text{secant-tangent theorem} \\ \mathbf{AC} & \mathbf{=} & \mathbf{ \frac{AD^2}{AB} }\\\\ CD^2 &=& AD^2 + AC^2 & \text{pythagoras} \\ CD^2 &=& AD^2 + AC^2 & | \quad AC= \frac{AD^2}{AB} \\ CD^2 &=& AD^2 + \left(\frac{AD^2}{AB}\right)^2 \\ CD^2 &=& AD^2 + \frac{AD^4}{AB^2} \\ CD^2 &=& AD^2\cdot \left( 1+\frac{AD^2}{AB^2} \right) \\ CD^2 &=& AD^2\cdot \left[1+\left(\frac{AD}{AB}\right)^2 \right] \\ \mathbf{CD} & \mathbf{=} & \mathbf{AD \cdot \sqrt{1+\left(\frac{AD}{AB}\right)^2 } }\\ \hline \end{array} \)
CD = ?
\(\begin{array}{|rcll|} \hline \mathbf{CD} & \mathbf{=} & \mathbf{AD \cdot \sqrt{1+\left(\frac{AD}{AB}\right)^2 } } \quad | \quad AB=4 \quad AD=8 \\ CD & = & 8 \cdot \sqrt{1+\left(\frac{8}{4}\right)^2 } \\ CD & = & 8 \cdot \sqrt{1+ 2^2 } \\ CD & = & 8 \cdot \sqrt{5 } \\ CD & = & 17.88854382 \\ \hline \end{array}\)
