This is a chapter out of my lecture notes in Geometry. Hopefully it will attract you to join our happy family of math students and faculty. We assume you know the basic properties of inverting the xy-plane in a given circle K.

Section 1. Harmonic Quadruples

First, we have to introduce the concepts of harmonic quadruples .

Def. Four points A,B,C,D form a harmonic quadruple (harm quad) if they are collinear and AC.BD = AD.BC. Then D is called a 4th harmonic point to A,B,C.

Cor. If (A,B,C,D) form a harm quad, then so do (B,A,C,D) and (A,B,D,C), (C,D,A,B), but not (A,C,B,D). Thus C is the 4th harmonic point to (A,B,D), A is the 4th harmonic point to (C,D,B) and to (D,C,B), B is the 4th harmonic point to (D,C,A), etc.
Proof. Just draw the diagrams on paper.QED

Constr 1.1. (Construction of 4th harmonic point by ruler alone. Diagram 1.1). Given 3 collinear points A,B,C. Construct the 4th harmonic point D by ruler only.
Soln. Choose any point E in the plane. Choose any point G on EB. Draw lines AG, CE, intersecting at K. Draw lines CG, AE, intersecting at F. Draw line KF, cutting line ABC at D.
Proof. By Menelaus' thm, (EK / KA) (AD / DC) (CF / FE) = 1.
By Cevas' thm, (EK / KA) (AB / BC) (CF / FE) = 1.
Therefore AD / DC = AB / BC. Thus D is 4th harmonic point to A,B,C. QED

Constr 1.2. (Construction of parallel line to 3 collinear points. Same diagram, with D at infinity, so that FK || AC). Given 3 collinear points A,B,C with B being mid point of AC. Given any point F in the plane not on the line ABC. Construct a line thru F || ABC.
Soln. Draw line FA. Choose any third point E on FA. Draw lines BE and CF intersecting at G. Draw lines AG and CE intersecting at K. Then line FK will be || ABC.
Proof. By Cevas' theorem, EF / FA = EK / KC . QED

Remark. The above 2 constructions form the basis of all constructions with ruler alone, as we shall see.

Section 2. Poles and Polars

Def. Given a circle K, center O. For any point P not = O, the set of points Q so that the line PQ would cut K at 2 points that form a harm quad with P and Q, is called the polar set of P w.r.t (with respect to) circle K.

Henceforth we assume a circle K with center O is given on the plane. Its radius will be taken to be 1 unit.

Thm 2.1. (The polar is a straight line. Diagram 2.1) For each point P not = O, the polar of P is a st.line thru the inverse P' of P, and perp. to OP. If P is on K, then the polar of P is the tangent line at P.

Proof. (Ingenious) Let PCD be a line thru P such that CD is a diameter of K. Let PAQB be any line thru P cutting K at A,B, and P,Q,A,B form a harm quad. Let lines BD cut AC at F and AD cut BC at E. Let lines FE cut PAB at Q' and PCD at P'. From Menelaus' thm, we have
(FB / BD) (DP / PC) (CA / AF) = 1. But from Cevas' thm we have
(FB / BD) (DP'/P'C) (CA / AF) = 1. Hence DP / PC = DP'/P'C, or C,D,P,P' form a harm quad. That means P' is the inverse of P in K.
Similarly, A,B,P,Q' form a harm quad, so that Q = Q'. Now E is the orthocenter of triangle CDF (and F is the orthocenter of triangle CDE), because angles CAD = CBD = rt.angle. Hence FP' is perp. to OP. Since Q is arbitrary, we have proven the polar set of P is a subset of the line through P' perp. to OP.
The above argument is valid whether P is inside or outside K. By a limit argument, if P moves towards K, so will P', so that the line FP' becomes the tangent at P when P is on K. If P is inside K, the polar set is the whole line FP', but if P is outside, the polar set is only a chord. QED

Def. The whole line L containing the polar set of P is called the polar of P, and P is calle the pole of the line L.

Thm 2.2. (One-one pairing between poles and polars.) Let L = all lines in the plane not thru O, P = all points in the plane not = O. Then the mapping f: P -> L with f(X) = polar of X for all points X in P, is a bijection.

Proof. Thm 2.1 proved that any point X has only one line as polar, hence f is a function.
If L1 and L2 are 2 lines in L, whether || each other or not, their feet Q1 and Q2 of perpendiculars from O will be different, hence the inverses Q1' and Q2' are different. Hence different lines L1, L2 have different poles Q1', Q2'. Hence f is injective.
Every line in L does have a pole, e.g., L1 has a pole Q1'. That proves f is surjective. QED

Thm 2.3. (I scratch your back, you scratch mine.) For any 2 points P,Q, if P lies on the polar of Q, then Q lies on the polar of P.

Proof. Case 1: If Q is inside the circle, then P is outside. Let line PQ cut K at A,B. Then (A,B,Q,P) form a harm quad. Therefore (A,B,P,Q) also form a harm quad. Hence Q lies on the polar of P.

Case 2: If P is inside the circle, then Q is outside, and we can argue similarly.
Case 3: Dia. 2.2. Finally, if both P and Q are outside K, then P lies on the polar of Q that cuts OQ orthogonally at the inverse Q' of Q. Let the segment Q'P cut K at A, then QA touches K at A, and since angle QAP > a rt.angle, the line QP does not cut K at all, so that N is outside, and N' inside K. From Case 1, with N' inside K, and Q lying on polar of N', therefore N' lies on the polar of Q, which is line PQ'. Now N' is the orthocenter of OPQ, hence line QN' cut OP orthogonally at some point P'. Since PP'N'N is cyclic quad, OP'.OP = ON'.ON, so that P' is inverse of P. Since line QP' cuts OP orthogonally, it is the polar of P, hence Q lies on the polar of P. QED

Cor. For each line L not thru O, the pole is the inverse of the foot of perpendicular from O. It also is the point cuncurrent to the polar of every point on L. For each point P not = O, its polar is not only the line thru its inverse P' and perpendicular to OP', it is also the congregate of the pole of every line through P.