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 Result
Nine Point Conic and Generalization of Euler Line.