AP*AP'*BQ*BQ'*CR*CR' = AR'*AR*CQ'*CQ*BP'*BP.

The relation is also sufficient i.e. if it is satisfied then there is a conic passing through the six points {P,P',Q,Q',R,R'}.

The relation follows by noticing that the ratios of the kind:

(AP*AP')/(AR*AR') = I(AC)/I(AB) (*).

Where I(AB) is a quadratic form depending only on the direction of line AB:

Here it is assumed that the conic is represented by an equation of the form:

Also line AB is assumed to be parameterized by the equation (variable r):

These relations were established in the discussion of the generalization of the power of a point with respect to a conic (See PowerGeneral.html ). They can be used here by applying (*) successively three times for the lines through {A,B,C} respectively:

(AP*AP')/(AR*AR') = I(AC)/I(AB),

(BQ*BQ')/(BP*BP') = I(AB)/I(BC),

(CR*CR')/(CQ*CQ') = I(BC)/I(AC).

The result follows by multiplying the three equalities and observing that the resulting right side equals to 1.

To prove the converse consider the conic passing through five of the six points and use the proven part and the valid relation to show that the sixth point is also on that conic.

and the relation becomes (simplifying squares and taking orientation into account).

AP*BQ*CR = - AR*CQ*BP.

This is Ceva's ( Ceva.html ) theorem and implies that

Assume now that

Hence, by the parallelity AQ passes through the middle of PR. By drawing a tangent B'C' parallel to BC we see that its middle Q' is also on line AQ.

The result is the well known property of conjugate diameters of conics.

Chords parallel to a given direction (BC) have their middles on another chord (QQ') passing through the contact points of BC and its parallel B'C'. Besides the tangents at the contact points (P,R) of the parallel chords intersect at points (A) lying on "conjugate" diameter QQ' of the conic.

(AP*AP')*(BQ*BQ')*(CR*CR')*(DS*DS') = (AS*AS')*(DR*DR')*(CQ*CQ')*(BP*BP').

Ceva.html

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