Consider a circle c(E,r) and a point A. On every secant line e = BA through A define the harmonic conjugate of A with respect to BC, i.e.

point D on e such that: DB/DC = - AB/AC. Then, for (e) rotating about A the corresponding point D moves on a line (f).

This is the

orthogonal to line EA.

Draw the circle through B, C and the center E of circle c(r). This intersects line AE at F. Triangle BCE is isosceles and this implies that

angle (BFA)=(ACE)=(ABE)=(AFC). Thus line AF is bisector of the angle BFC. Hence point D being harmonic conjugate to A with

respect to {B,C} is on the outer bisector and angle DFA is a right one.

Let IJ be the diameter on AE. Line BI bisects angle ABF. This is seen by first noticing the similarity of triangles BFE and BAE, due

again to the isosceles property of BEC. In fact (BFI)+(FBI)=(BIE) and (IBA)+(ABE)=(IBE)=(BIE). Using (BFI)=(ABE) =>

(FBI)=(IBA) and since IBJ is a right angled triangle points {I,J,A,F} build a harmonic division i.e. F belongs to the polar of A. This

proves all claims.

conjugate with respect to {A,D}. This is a symmetric relation and implies the so-called

If the polar p

In other words this means that all the polars p

1) AE*FE = r

which is the inverse of A with respect to c(r).

2) The inverse of the polar (f) with respect to c(E,r) is the circle (d) with diameter AE.

3) Every point K on the the polar (f) is inverted into some point H and by (1) line AH is the polar of I.

4) The polar (f) of point A is the inverse of the circle (d) with diameter AE.

5) Quadrangle KCEF being cyclic, the tangents CK, KB of c(E,r) at B and C meet at the polar of A.

6) For points A lying exactly on the circle c(E,r) the corresponding polar is the tangent at A.

7) By convention the polar of the center E is the line at infinity. Inversely every point at infinity A* determines a

direction (v) of parallel lines and the polar of A* is identified with the line through the center E whose direction

is orthogonal to v. For a practical construction of the polar of a point see the file Polar2.html . Some further properties of the polar are

discussed in the file Polar3.html and Polar4.html .

lines {AX, AY, BC} respectively at points {H,F,G} such that {E,G} are harmonic conjugate with respect to {F,H}.

Let I be the intersection point of AY with BE which is the polar of H. Then the polar p

H by the pol-polar reciprocity and passes obviously through C. This, using the harmonic pencil at H: H(I,C,A,F), implies that

{A,F} are harmonic conjugate to {C,I}. Hence the pencil of lines at B: B(A,F,C,I) is also harmonic, hence its intersection

with line t

Given a conic (c) and a point A the polar p

D of A with respect to {B,C}, where later are the intersection point of the conic with secants through A.

The fact that the geometric locus described above is a line, can be seen in many ways. Two particularly easy are:

1) By deriving it through a projectivity from the corresponding property of the circle.

2) By using the equation of the conic (see Conic_Equation.html , section-5).

The figure below illustrates the usual way to draw the polar p

Draw two secants {BC, B'C'} through A and find the intersection points {E,E'} of opposite side-pairs (BC',CB') and (BB',CC').

Line EE' is the polar of A.

The reason for this can be drawn back to Pascal's theorem (see Pascal.html or even better PascalOnQuadrangles.html ) by which the

tangents at {B',C'} concure also at a point F of line EE'. Then turning the secant B'C' about A we move D' on line p

{B',C',A,D'} build a harmonic division. Line p

Polar2.html

Polar3.html

Polar4.html

Conic_Equation.html

Pascal.html

PascalOnQuadrangles.html

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