# Can someone please summarize or explain this?

Theorem 120

If a tangent and a secant segment intersect in the exterior of a circle, then the square of the length of the tangent segment is equal to the product of the lengths of the entire secant segment and its external part

### 1 Answer

- SamwiseLv 72 years agoFavorite Answer
Draw yourself a circle, and mark a point P some distance outside it.

From P, draw a line tangent to the circle at a point labeled T.

(By "tangent to the circle," we mean that it touches the circle only at point T,

rather than going through any of the circle's interior.)

From P, draw another line that intersects the circle,

this time going through part of it, so it intersects the circle at two points.

(That's what's meant by a "secant.")

The intersection point near P we'll call N,

and the intersection point further away from P we'll call F.

Now, instead of the whole lines, we'll focus on the line segments with named endpoints:

PT is the tangent segment, and

PF is the entire secant segment,

with PN the "external" part of the secant segment (that is, the part outside the circle).

Since we're interested in the lengths of segments, we can use the notation where

|AB| would represent the length of segment AB.

The theorem tells us that for our figure,

|PT| squared = the product of |PF| and |PN|.

Thank you so much!!