Theorem. If the rectangle (AP.BP) contained by the segments of a secant, drawn from any point (P) without a circle, be equal to the square of a line (PT) drawn from the same point to meet the circle, the line which meets the circle is a tangent.
Dem.—From P draw PQ touching the circle [xvii.]. Let O be the centre. Join OP, OQ, OT. Now the rectangle AP.BP is equal to the square on PT (hyp.), and equal to the square on PQ [xxxvi.]. Hence PT2 is equal to PQ2, and therefore PT is equal to PQ. Again, the triangles OTP, OQP have the side OT equal OQ, TP equal QP, and the base OP common; hence [I. viii.] the angle OTP is equal to OQP; but OQP is a right angle, since PQ is a tangent [xviii.]; hence OTP is right, and therefore [xvi.] PT is a tangent.