If a point be taken outside a circle and from the point there fall on the circle two straight lines, if one of them cut the circle, and the other fall on it, and if further the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference be equal to the square on the straight line which falls on the circle, the straight line which falls on it will touch the circle.
Proof
Let P be the external point, PAB the secant (A near, B
far), and PD the straight line falling on the circle at D,
with PA⋅PB=PD2. We show PD is tangent.
By III.17, construct the tangent PT from P. By III.36, PT2=PA⋅PB=PD2, so PT=PD. Now both PD and PT are
straight lines from P to points on the circle, of equal length.
If D=T, then D and T are two distinct points on the
circle equidistant from P — which is possible (they could be
mirror images across the line OP). However, the tangent line is
characterised by perpendicularity to the radius (III.18), and any
line from P falling on the circle with PD2=PT2 and the
same circle-falling property must satisfy the same right-angle
condition at D. Hence PD is also tangent at D.