1) The tangents at opposite vertices of q intersect at two points I, J lying on the polar e of the intersection point of the diagonals E.

2) The bisectors g, h, of the angles formed by the opposite sides of q intersect at a point M on the circle with diameter GH, where G, H are the intersection points of the opposite sides of the quadrilateral.

3) (I,J,G,H) = -1 i.e. the four points build a harmonic tetrade and lines g, h are also the bisectors of the angle at M of triangle IMJ.

1) follows from the duality of pole-polar. Since E is on the polar of I, I must be also on the polar of E which is e. 2) follows by measuring the angle of triangle HGM at M in terms of the angles of q. 3) is a bit more complicated involving Brianchon's theorem, by which the diagonals of the circumscribed quadrilateral q' = NOPQ, build from the tangents at the vertices of q intersect also at point E. Then, by the basic figure discussed in Harmonic.html , follows that (I,J,G,H) = -1 and this implies the assertion on the bissectors of triangle IMJ, becasuse of the orthogonality of g, h.

There is an elementary proof of Brinachon's theorem (for quadrilaterals) discussed in the document: CircumscriptibleQuadrilateral.html .

Harmonic bundles having two of the lines orthogonal, which bisect the angle of the other two, are discussed in Harmonic_Bundle.html .

Produced with EucliDraw© |