Explore
Construct the orthocentres E, F, G & H, respectively of triangles ABC, BCD, CDA & DAB, of any quadrilateral ABCD.
1) Drag any of the vertices of quadrilateral ABCD.
What do you notice about the areas of the two quadrilaterals? Is the result still valid if ABCD is concave or crossed?
2) Click on the 'Link to cyclic quad' button at the bottom to navigate to the case when ABCD is cyclic.
Click on the given 'Show Measurements' button in this sketch to display corresponding side & angle measurements, as well as the 'Show Objects' button to show the lines connecting corresponding vertices.
What do you notice about the relationship between the two quadrilaterals? Is the result still valid if ABCD is crossed?

#### .sketch_canvas { border: medium solid lightgray; display: inline-block; } The Orthocentre Quadrilateral of a Quadrilateral

Challenges
1) Can you explain why (prove that) the two areas of the two quadrilaterals above remain equal to each other?
2) Can you explain why (prove that) in the special case when ABCD is cyclic that FGHE is congruent (and corresponding sides parallel) to ABCD? Or alternatively stated, explain why (prove that) ABCD can be mapped on to FGHE with a half-turn (around the 'anticentre' (nine point or Euler centre) of cyclic ABCD)?

Notes
1) This very interesting result in 1) above is stated (on p. 475) without proof as an empirically confirmed conjecture by Maria Mammana & Biagio Micale in their paper Quadrilaterals of triangle centres in the Mathematical Gazette, Vol. 92, No. 525 (November 2008), pp. 466-475.
2) Mammana & Micale (2008) provide a proof of the 2nd result above when ABCD is cyclic (in their Theorem 2). Their excellent paper also proves several other interesting results about quadrilaterals formed by the triangle centres of the triangles subdividing a quadrilateral, and is definitely worth a read.

Acknowledgement
I'm grateful to Vladimir Dubrovsky, Kolmogorov School of Moscow State University, The Russian Federation, for bringing this interesting & challenging problem to my attention.

Proof
On 20 July 2022, Istvan Biro from Romania produced the following proof in 'Training Geometry Olympiad' Group on Facebook using complex numbers & computer algebra software to confirm the result - read it at: Biro-proof.

Related Result
Of possible interest to some readers might also be the following result about the quadrilateral formed by the circumcentres of triangles ABC, BCD, CDA & DAB of a tangential quadrilateral ABCD, as well as the repetition of the same construction on the newly formed quadrilateral - go here.

By Michael de Villiers. Created with WebSketchpad, 14 July 2022.