Construction:
• Construct line OP intersecting gamma
at points A and B such that A lies between O and P.
• Construct a secant through P intersecting gamma at points
C and D such that D lies between C and P.
• Construct lines AC and BD intersecting at point E.
• Construct lines AD and BC intersecting at point F.
• Construct line EF intersecting gamma at point G such that
E lies between G and F.
• Construct the tangent line PG.
Proof:
We must prove that OGP is a right angle.
Label the intersection of lines EF and OP as H.
Note that ABCD and CFDE are cyclic quadrilaterals.
In the following three letters denote an angle unless otherwise specified.
Let BAC = x and ABD = y. It is then easy to show that
BDC = x
CDF = ADP = 90  x
CEF = HEA = 90  x
AHE = 90
ACD = y
APC = x  y
In the following let r = OA = OB = OG
From right triangles AHE and BHE,
(r  OH)tan(x) = (r + OH)tan(y) (1)
Applying the law of sines to triangle ADP,
AD AD AP OP  r
 =  =  =  (2)
sin(x  y) sin(APD) sin(ADP) sin(90  x)
From right triangle ADB,
AD = 2r sin(y) (3)
Combining equations (1)(3) gives,
OHOP = r^{2}
From right triangles OHG and PHG,
OG^{2}  OH^{2} = PG^{2}  PH^{2}
= PG^{2}  (OP  OH)^{2}
= PG^{2}  OP^{2} + 2OHOP  OH^{2}
= PG^{2}  OP^{2} + 2OG^{2}  OH^{2}
or
OP^{2} = PG^{2} + OG^{2}
Therefore, OGP is a right angle.
Note: The secant PAB does not have to go through the center O for the construction to work, but I don't have a simple proof if it doesn't.
Note: The line EF is said to be the polar line of the point P with respect to gamma and the point P is said to be the pole of line EF.
Note: The points P and H are said to be inverse points with respect to gamma.
