Mathot2

Mathot point of a circular quadrangle (2)

Continuing on Mathot.html . Draw the orthocenter P of the triangle ABC, which is part of the circular quadrangle ABCD. CP is orthogonal to AB hence, drawing from G parallel to CP we get GM going through the Mathot point. The same is true for HM, hence their intersection point coincides with the Mathot point of the quadrangle. This gives the following proposition:
The quadrangle of the orthocenters of the 4 partial triangles of a circular quadrangle form a quadrangle which is symmetric to the original, with respect to the Mathot point of the original quadrangle. By symmetry the Mathot point of the second quadrangle will coincide with the Mathot point of the original.

Look at the document: Mathot3.html for the complete picture, including the two symmetric quadrangles.

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