For every circumscriptible pentagon the line (AL, look at [concurring-1]) joining a vertex to the contact point of the opposite side and the two diagonals (CE and DB ) from the end points of that side to the neighbour vertices of the initial vertex (A) intersect at a point (U).

The diagonals and the lines joining opposite contact-points, in a circumscriptible to a circle quadrilateral, pass all through a common point O. The proof can be given by specializing Brianchon's theorem, explained in Brianchon.html .

There is also another more elementary proof discussed in the document: CircumscriptibleQuadrilateral.html , whereas CircumscriptibleQuadrilateral2.html contains an alternative proof and further properties of this figure.

Corresponding theorems are valid also for hexagons/pentagons/quadrilaterals circumscriptible to conics: Brianchon3.html .

Brianchon3.html

Circumscriptible.html

CircumscriptibleQuadrilateral.html

CircumscriptibleQuadrilateral2.html

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