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This essential trains for: AMC-10, AMC-12, AIME.

The product of the segments determined by the intersections of a circle and a ray is invariant with respect to the direction of the ray and equal to the square of the length of the tangent to the circle from the origin of the ray.

Given a point P outside a circle of center O, there are two lines through P that are tangent to the circle. The lengths of the two segments determined by P and the points of contact are equal (common tangent theorem).

Denote one such point of contact with T.

An arbitrary ray from P intersects the circle at A and at B.

Then:

Since the ray was arbitrary, the same equality works for any other such ray:

Which is equivalent to the proportionality relationship:

Proof: To prove a proportionality relationship among segments it is a good idea to look for similar triangles.

To identify similar triangles in the figure, congruent angles are a good clue:

Inscribed angles as well as angles formed by a tangent and a chord are equal in measure to the half of the intercepted arc. Therefore:

The angle atP is common to both triangles PTB and PAT.

Since the angles of a triangle add up to 180°, the angles PTB and PAT must be congruent.

The triangles PTB and PAT have the same angles and are, therefore similar. Hence, the similarity ratios among side lengths:

where from, by cross-multiplying:

If the point P is interior to the circle the proof is much more direct, since the triangles formed are similar in a more obvious way.

Since the angles ∠MPA and ∠BPN are vertical angles, they are congruent.

Also, since ANBM is a cyclic quadrilateral, the angles ∠MAP and ∠BNP are congruent.

Therefore, the triangles MAP and BNP are similar and the ratios between corresponding side lengths are: