# What is the measure of the secant CD?

The measure of secant CD is 7.

I arrived at this surprising result by making use of the Pythagorean theorem in the following ways. Define the following points:
S is the midpoint of secant AB
T is the midpoint of secant CD
P is the center of the circle.
The following relations apply. These make use of the Pythagorean theorem.
(PS)^2 + (SB)^2 = (PB)^2 = (PD)^2    (PB and PD are both equal to the radius of the circle)
(PT)^2 + (TD)^2 = (PD)^2
(PS)^2 + (SB + 2)^2 = (PX)^2
(PT)^2 + (TD + 3)^2 = (PX)^2
We can equate the first two equations and we can equate the last two equations. Doing this gives
(PS)^2 + (SB)^2 = (PT)^2 + (TD)^2
(PS)^2 + (SB + 2)^2 = (PT)^2 + (TD + 3)^2
Subtracting the first of these two equations from the second gives
(SB+2)^2 - (SB)^2 = (TD+3)^2 - (TD)^2
4(SB) + 4 = 6(TD) + 9    (simplify)
4(13/2) + 4 - 9 = 6(TD)    (substitute the value of SB, which is half of 13; subtract 9)
21 = 6(TD)    (collect the constants)
7 = 2(TD) = CD    (divide by 3 to get the value of 2(TD) = CD)

A Google search of "intersecting secants" produces a theorem that covers this very simply.

It says the product of BX and AX is equal to the product of CX and DX. The former is 2*15 = 30. In order for the latter product to be 30, CX must be 30/3=10. Thus CD must be 10-3 = 7.
thanked the writer.