[alogo] Orthodiagonal from cyclic

For every cyclic quadrilateral q = ABCD, there is associated an [orthodiagonal] quadrilateral p = KLMN with the following properties:
1] p is circumscribed about q.
2] The diagonals LN and KM of p pass from points F and E, intersection points of side-pairs (AD,BC) and (AB,CD) respectively.
3] The intersection point O of the diagonals LN and KM is on the perpendicular from H to line EF, H being the intersection point of the diagonals of q.
4] Lines OA, OB, OC and OD are perpendicular to the corresponding sides of p.
5] The other points of intersection of these lines with p lie on the circumcircle c of q and build a rectangle r = A*B*C*D*.

[0_0] [0_1]
[1_0] [1_1]

It is known that H is the pole of EF (look at CyclicProjective.html ) and the circle (d) with diameter EF is orthogonal to (c). Draw the HG perpendicular to EF and define its intersection point O with the circle (d), on same side with the center I of (c). Draw then orthogonals to OA at A, OB at B, ... etc. defining the quadrangle p = KLMN. The quadrilaterals q1 = OAKD and q2 = OBMC are cyclic and F is on their radical axis, since FAD, FBC are chords of the circumcircle (c). It is easy to see that |FO| = |FP||FQ| = |FA||FD| = |FB||FC|. This implies that FO is orthogonal to the diameters KO, OM of circumcircles of q1 and q2, i.e. the circumcircles of q1 and q2 are tangent to line FO at O. Hence K, O, M are on a line and FO is perpendicular to this line, hence KM passes through E. Analogously one shows that LN passes through F. The last (5th) assertion is a general property of orthodiagonal quadrangles.

The construction of the figure here is, in some sense, the inverse of the procedure explained in the file Orthodiagonal.html . See this file for a proof of the 5th assertion.

Equality |FO| = |FP||FQ| is a consequence of the orthogonality of two circle bundles handled in OrthogonalBundles.html .

Some additional facts on these kind of quadrilaterals are discussed in the file Orthogonal_Diagonals.html .

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