1) The diagonals AC, BD of the quadrilateral intersect at a point E, which is the pole with respect to the circumcircle (b) of the line (d), (d) being the (red) line through the intersection points F, G of opposite sides of q.

2) The diagonals HK, IJ, of the quadrilateral q' of contact points of q with the incircle, intersect also at E orthogonally.

3) The polar of E with respect to the two circles is line (d).

4) Circles (a), (b) generate a (coaxal) circle bundle (I) of non intersecting type. The member circles of this bundle are invariant under the projective collinearity T of the quadrangle q. T is identical with the projective collinearity T' of the quadrangle q'.

5) The radical axis of the two circles (a) and (b) is line g, parallel to (d) at half distance from E.

6) The incenter, the circumcenter and the intersection points of the diagonals of q are on a line (f).

7) The circle with diameter FG is orthogonal to the circumcircle and passes through the incenter of q.

The proofs follow from the remarks:

- (1) is a general property of cyclic quadrilaterals (see the file CyclicProjective.html ). The same reference proves the orthogonality of the cirlce (e) to the circumcircle (b) of q.

- (2) follows from Brianchon's theorem. Besides, from (1) follows that (K,H,E,M) = (J,I,E,L) = -1 i.e. the polar line of E with respect to circle (b) is also polar of E with respect to the incircle (a). This proves (3). The triangle KHF being isosceles its basis KH is orthogonal to FY and parallel to the side YG of the right angled triangle FYG.

- (3) Since the two circles are separated by the quadrangle q, the bundle I(a,b) generated by (a) and (b) is of non intersecting type. Consider now the projective collinearity T of q. By its definition it interchanges A, C and B, D. In particular maps Triangle ADC to CAB and preserves its circumcircle. By the general properties of this kind of transformations it has as fixed points the points of line (d) together with the point E. Thus it coincides with the projectivity leaving invariant the circle (b) and mapping points A, D, C to C, A, B respectively. By the general properties of these projectivities T leaves every circle of the bundle generated by (b) and the point E. Considering the projective collinearity T' interchanging H, K and I, J we get a similar result for the bundle generated by (a) and E. But T, T' coincide on H, K, I and J, hence they coincide everywhere. This proves (4).

- By the characteristic properties of harmonic tetrads of points, we have from (Q,P, O,E) = -1, that the middle S of OE satisfies OE^2 = OQ*OP and a similar relation for the diametral points on (a) defined by line f. Thus S belongs to the radical axis of the two circles. This proves (5).

- Consider now the segments GY, FY, Y being the incenter and measure their angle. It is easily proved to be pi/2. Thus point Y belongs to the circle (e). This completes the proof of all statements.

Look at the file ProjectiveCollinearityQuadrangle.html for a discussion of the [Projective Collinearity].

For the various forms of Brianchon's theorem see the file Brianchon2.html . There is given a reference to an elementary proof for the case of a quadrangle.

For the definition and discussion of properties of harmonic tetrads of points see the file Harmonic.html . Notice that EucliDraw has a tool constructing directly a bicentric quadrilateral (called there [Bicircular]). The figure above shows also a point X and its image T(X) as well as the point X' on (d) cut out by the line [X,T(X)]. By the general properties of the projectivities of this kind the tetrad is harmonic (X,T(X),E,X') = -1.

The file Bicentric2.html discusses the images under T of the circles-members of the orthogonal bundle to I(a,b).

The file Orthocycle.html initializes a discussion leading to a classification of all the bicentric quadrilaterals inscribed in a given circle.

The file Poncelet.html gives a condition for the radii of inscribed and circumscribed circles of a bicentric quadrilateral.

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