Pascal in a Cyclic Quadrilateral: What is this about?
A Mathematical Droodle

Explanation

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A nice problem has been posted by Darij Grinberg at the Hyacinthos group:

An elegant solution was found by Jean-Pierre Ehrmann.

The lines perpendicular to AC and BD form a parallelogram whose center coincides with the center O of the circle. Therefore E, F, and O are collinear. (In fact O is the midpoint of EF.)

In order to prove that X lies on the same line, construct the antipodes A' of A and D' of D on the circle. Note that Pascal's theorem is applicable to hexagon ABD'DCA'. The theorem says that the three points E, F, and X lie on Pascal's line of the hexagon.

Pascal and Brianchon Theorems

• Pascal's Theorem
• Pascal in Ellipse
• Pascal's Theorem, Homogeneous Coordinates
• Projective Proof of Pascal's Theorem
• Pascal Lines: Steiner and Kirkman Theorems
• Brianchon's theorem
• Brianchon in Ellipse
• The Mirror Property of Altitudes via Pascal's Hexagram
• Pappus' Theorem
• Pencils of Cubics
• Three Tangents, Three Chords in Ellipse
• MacLaurin's Construction of Conics
• Pascal in a Cyclic Quadrilateral
• Parallel Chords
• Parallel Chords in Ellipse
• Construction of Conics from Pascal's Theorem
• Pascal: Necessary and Sufficient
• Diameters and Chords
• Chasing Angles in Pascal's Hexagon

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