BundleQuadrangle

Bundle quadrilaterals

Consider a circle bundle of non intersecting type and two chords of a member circle passing through the limit point of the bundle. The chords define a quadrilateral q = ABCD having these as diagonals. Extend two opposite sides AD, BC until they intersect a second circle member of the bundle. The intersection points build a quadrilateral r = HIJK. Show that the intersection point of the sides HI, JK lies on the polar MN of E w.r. to a circle of the bundle (all circles c of the bundle have the same polar with respect to E).

Corollary: The diagonals of all quadrilaterals HIJK intersect at E.

In fact, N, M can be taken as the intersection points of opposite sides of q. Then N is on the polar of E, hence the polar p(N) of N contains E. Consider the intersection points O, P of this polar with sides HK, IJ respectively. Assume also L to be the intersection point of HI, DC. Then a) these sides intersect at a point L lying on p(N). b) L is on line MN. a) follows from the standard theorem on cyclic quadrilaterals (see CyclicProjective.html ). b) follows from the fact that the quadrupple of lines at N (NL,NH,NE,NI) is harmonic. But (NM,NH,NE,NI) is also harmonic, hence L is contained in line MN.
The corollary follows also from the fact that for cyclic quadrilaterals q=ABCD the diagonals intersect at the pole E of MN (see reference above).

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