When two secant lines intersect each other outside a circle, the products of their segments are equal.

(Note: Each segment is measured from the outside point)

Try this
In the figure below, drag the orange dots around to reposition the secant lines.
You can see from the calculations that the two products are always the same.
(Note: Because the lengths are rounded to one decimal place for clarity, the calculations may come out slightly differently on your calculator.)

This theorem works like this: If you have a point outside a circle and draw two secant lines (PAB, PCD) from it, there
is a relationship between the line segments formed. Refer to the figure above. If you multiply the length of PA
by the length of PB, you will get the same result as when you do the same thing to the other secant line.

More formally: When two secant lines AB and CD intersect outside the circle at a point P, then

PA.PB = PC.PD

It is important to get the line segments right. The four segments we are talking about here all start at P, and some overlap each other
along part of their length; PA overlaps PB, and PC overlaps PD.

Relationship to Tangent-Secant Theorem

In the figure above, drag point B around the top until it meets point A. The line is now a tangent to the circle, and PA=PB.
Since PA=PB, then their product is equal to PA^{2}. So:

PA^{2} = PC.PD

This is the Tangent-Secant Theorem.

Relationship to Tangent Theorems

If you move point B around until it overlaps A, the resulting tangent has a length equal to PA^{2}. Similarly,
if you drag D around the bottom to point C, the that tangent has a length of PC^{2}. From the this theorem

PA^{2} = PC^{2}

By taking the square root of each side:

PA = PC

confirming that the two tangents froma point to a circle are always equal.