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

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**.