Six Point Cevian Circle

Exploration
As is well-known, the famous nine-point circle passes through the midpoints of the sides (feet of the medians) of the associated triangle as well as through the feet of the altitudes. It therefore seems natural to wonder what happens if instead of the medians, one started with an arbitrary set of concurrent cevians AD, BE & CF as shown below. Then carry out the instructions as follows:
Step 1: Click on the 'Show Circumcircle DEF' button.
Step 2: Click on the 'Show cevians 2' button.
1) What do you notice about the 2nd set of cevians drawn to the other intersections of circle DEF with the sides of the triangle?
2) Formulate your findings & explore further by dragging.
Note: Drag D past G or E past H, or onto the extensions of the sides.

Six Point Cevian Circle

Formulation
Given any ∆ABC and a set of concurrent cevians AD, BE & CF, then the cevians AG, BH & CI (drawn respectively, from A, B & C to the other intersections G, H & I of the circumcircle DEF with sides BC, CA & AB), are also concurrent.
Note 1: This result is apparently also known as the Cyclocevian Conjugate theorem on Wolfram MathWorld, as well as Terquem's Theorem on Cut the Knot and Reuschle's theorem on Wikipedia.
Note 2: The result is not only a generalization of the nine-point circle, but also of the Gergonne point of a triangle. Click on the 'Link to Incircle Special Case' button. In the new sketch drag the relevant points until the circle DEF coincides with the incircle. When that happens, points X and Y will coincide with the Gergonne point.

Challenge
Can you explain why (prove that) the result above is true?
Hint: Try using Ceva's theorem, then click on the 'Link to Six Point Circle Proof' button to check or compare your proof.

Conic Generalization & more
a) If two arbitrary sets of concurrent cevians (e.g. AD, BE & CF, and AG, BH & CI) are constructed in any ∆ABC, then their feet (e.g. DEHFIG) lie on a conic (and is called the bicevian conic of X and Y).
Click on the 'Link to Generalization' button to view a dynamic sketch of the generalization in a) above.
b) Click on the 'Show More Properties' button & observe that the diagonals of the formed hexagon LMNOPQ are concurrent. Therefore, according to the converse of Brianchon's theorem this also implies that LMNOPQ has an inscribed conic, but is not shown.
Challenge: Can you prove the results above?
Hint: Try using the theorems of Pappus (and its dual), and that of Pascal. Alternatively, use Ceva's theorem together with Carnot's Conic Theorem.

Related Links
Nine Point Conic and Generalization of Euler Line
Euler line proof
Homothetic Polygons Concurrency
The Center of Gravity (Centroid) of a Triangle (Rethinking Proof activity, incl. Ceva)
Experimentally Finding the Centroid of a Triangle with Different Weights at the Vertices (Ceva's theorem)
Triangle Altitudes (Concurrency, orthocentre, Rethinking Proof activity)
Further Euler line generalization
Spieker Conic and generalization of Nagel line
Euler and Nagel lines for Cyclic and Circumscribed Quadrilaterals
Nine-point centre (anticentre or Euler centre) & Maltitudes of Cyclic Quadrilateral
Generalizing the Nagel line to Circumscribed Polygons by Analogy
The quasi-Euler line of a quadrilateral and a hexagon
Concurrency and Euler line locus result
Eight Point Conic for Cyclic Quadrilateral
Rigby's Eight Point Conic for a Quadrilateral
The affine invariance of the conics
British Mathematical Olympiad problem: Conic Generalization
Parallel-Hexagon Concurrency Theorem
Merry Go Round the Triangle
An interesting collinearity
The quasi-circumcentre and quasi-incentre of a quadrilateral

External Links
Ceva's theorem (Wikipedia)
Pappus (Wikipedia)
Pascal (Wikipedia)
Carnot's Conic Theorem (Wikipedia)
Aimssec Lesson Activities (African Institute for Mathematical Sciences Schools Enrichment Centre)
UCT Mathematics Competition Training Material
SA Mathematics Olympiad Questions and worked solutions for past South African Mathematics Olympiad papers can be found at this link.
(Note, however, that prospective users will need to register and log in to be able to view past papers and solutions.)

***************

Free Download of Geometer's Sketchpad & Associated Learning/Instructional Modules on Various Topics