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Pascal in a Cyclic Quadrilateral: What is this about?
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A nice problem has been posted by Darij Grinberg at the Hyacinthos group:
| Given a cyclic quadrilateral ABCD with the circumcenter O. The perpendicular to BD through B meets the perpendicular to AC through C at E. The perpendicular to BD through D meets the perpendicular to AC through A at F. Finally, let X be the intersection of the lines AB and CD. Then, the points O, E, F, X are collinear. |
An elegant solution was found by Jean-Pierre Ehrmann.
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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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