**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?

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.

*Back
to "Dynamic Geometry Sketches"*

*Back
to "Student Explorations"*

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