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1) Given a quadrilateral *ABCD* as shown below. Let *K*, *L*, *M*, and *N* be the respective circumcentres of triangles *ABD*, *ADC*, *BCD* and *ABC*, what do you notice about the intersection *O* of *KM* and *LN*?**Challenge 1**: Can you explain why (prove) your observation is true?

Quasi-circumcentre of quadrilateral

You should have noticed that the intersection *O* of *KM* and *LN* is equidistant from opposite vertices *A* and *C*, as well as equidistant from opposite vertices *B* and *D*. We shall call point *O* the *quasi-circumcentre* of *ABCD*.

**Renate's Theorem**: The above result about the quasi-circumcentre was experimentally discovered and proved from a problem posed in [1] to find the “best” place to build a water reservoir for four villages of more or less equal size, if the four villages are not concyclic. It followed from the classroom discussion of a proposed solution by an undergraduate student, Renate Lebleu Davis, at Kennesaw State University during 2006.

This result was used in the Kennesaw State Mathematics Competition for High School students in 2007, as well as in the World InterCity Mathematics Competition for Junior High School students in 2009.

**Reference**: [1] M. de Villiers, *Rethinking Proof with Sketchpad*, Emeryville: Key Curriculum Press, 1999/2003.

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2) Given a quadrilateral *ABCD* as shown below. Construct the angle bisectors for each of the four angles. Label *E* the intersection of the angle bisectors of angles *A* and *B*, label *F* the intersection of the angle bisectors of angles *B* and *C*, label *G* the intersection of the angle bisectors of angles *C* and *D*, and label *H* the intersection of the angle bisectors of angles *D* and *A*. What do you notice about *I* be the intersection of *EG* and *FH*?

**Challenge 2**: Can you explain why (prove) your observation is true?

Quasi-incentre of quadrilateral

You should have noticed that *I* the intersection of *EG* and *FH*, is equidistant from opposite sides *AD* and *BC*, as well as equidistant from opposite sides *AB* and *CD*. We shall call the point *I* the *quasi-incentre* of *ABCD*.

**Comment**: Note that both above results remain valid if *ABCD* is concave or crossed - check by dragging!

**Further generalization & application**: The quasi-circumcentre can be extended to a quadrilateral and a hexagon where it is collinear with the so-called *quasi-orthocentre* of a quadrilateral and *centre of gravity*, lying on the so-called quasi-Euler line of a quadrilateral and a hexagon.

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By Michael de Villiers. Created, 23 November 2014.