Another interesting configuration, this time associated with a more tl;dr post. Firstly, a complete quadrilateral is the figure formed by four lines intersecting at points, and the difference with a complete quadrangle is that this thing has three diagonals.
Before starting, I’d like to present a set of disclaimers.
1. This post is around 40% compilation and 60% observations. I have used geogebra for experimenting with the different configurations, and also searched a lot on the internet for different properties of this configuration. In the meanwhile, there are a lot of properties that I have independently figured out, but some I cannot prove myself.
2. I have mentioned “Awaiting proofs” to whichever property I have not been able to solve. For example, in Property 13, I was not able to prove all of the results thanks to the complexity. If I am able to come up with solutions, I will add them later. However, anyone eager to try is free to try and prove these results. Discussions regarding these are welcome as well.
3. Nothing else. Go ahead and enjoy reading. 🙂
Whenever we talk about complete quadrilaterals, we firstly think of Miquel’s theorem. A really well-known and useful theorem as it is, we will check it out along with some other basic results. I will refer to the following configuration always, ’cause it’s annoying to name each of the points every time I state a new theorem/result. We will just analyse the different properties of this configutation:
Let the lines meet in six points (named or unnamed 😛 ). Then the circumcircles of the four triangles that are formed out of these six vertices concur at a point, known as the Miquel Point, which I will designate as
Surely this is well-known and simple angle-chasing. I won’t give a proof for this one, since the post is getting more and more tl;dr everyday!
Property 2.(Miquel Circle Theorem)
The circumcentres of these four triangles formed are concyclic and lie on a circle that passes through the Miquel Point.
Refer to the diagram in Property 5. This Miquel circle doesn’t really look too hard, just note that Using this, working backwards of PP5, it is obvious that are concyclic. So, we are done.
Property 3.(Miquel-Steiner Line)
The orthocentres of these triangles lie on a line, known as the Miquel-Steiner Line.
For the proof of this, we make use of PP12. So maybe you should first scroll down a little, read what it is all about, then come back. 🙂
Obviously the midpoints of lie on the pedal line mentioned in PP12, and therefore their image after homothety wrt lie on a line. This line is known as the orthocentric or Miquel-Steiner line.
The orthocentre of lies on
I don’t really care about the configurations much, so I’d like you to refer to the diagram in the next Property. Now, according to the notations of PP5, we need to show that
Note that because the reflection of in lies on the Miquel Circle.
is the centre of spiral similarity that maps to
Property 6.(Newton-Gauss Line)
The midpoints of are collinear.
Not really too eager to write it all down. The following stronger version of this theorem holds, which is known as Gauss-Bodenmiller theorem.
Extension of property 6:
The circles drawn on the diagonals of a complete quadrilateral taking them as diameters, are coaxal.
For a proof, see here.
Define the unique point on such that and similarly define Then let be the midpoints of Then the points and lie on a straight line.
Note that the sides are harmonically separated by So, and are harmonically divided by and Since is a parallelogram, so the point is the point at infinity of Therefore So, we are done.
The projections of Miquel point onto the sidelines lie on a line perpendicular to the Newton-Gauss line.
Let the points be defined as in our previous property. Then it can be seen that are concyclic. So, is the Miquel point of both and
Using this, we see that projection of on is collinear with projections of on This line is the pedal line of The fact that this is orthogonal with the Newton-Gauss line also immediately follows.
Using this property, it can be shown that the Miquel point lies on the Nine Point Circle of the triangle where and This triangle is the diagonal triangle of
We define the Hewer Point as the centre of the circle that passes through the orthocentres of . Thus this point is the pseudo-orthocentre of
Since the point is defined as the circumcentre of therefore it is forced that the circle is congruent to the circle and the points correspond to the vertices meaning that the segments are bisected at a common point, so that is a parallelogram. Therefore, This gives us where the last line follows from in any triangle So this establishes that is the pseudo-orthocentre of
If is the Hewer point, the centre of the Miquel circle, the orthocentre of then we have and form a parallelogram.
Since is the pseudo-orthocentre of we already have that
And all these similar properties give us our desired results.
If the Euler lines of the four triangles formed, ie meet the Miquel circle again at points then and similars.
Due to the spiral similarity that maps and to and note that we have and also which leads us to the fact that and are isogonal conjugates wrt Because passes through the circumcentre, so its isogonal passes through the orthocentre of In other words, leading to the fact that is the point of concurrence of the four altitudes of through the vertices
DIAGRAM TO BE ADDED HERE.
For an arbitrary quadrilateral the nine-point circles of triangles and the pedal circles of the corresponding remaining point, are concurrent.
The proof makes use of a known property of conics, ie the pedal circle of a fixed point chosen on a conic wrt all the triangles inscribed in it pass through a fixed point. For rectangular hyperbola, this fixed point coincides with the centre. Thus every conic through the orthocentre and vertices of a triangle is a rectangular hyperbola, and because the centres of all these conics lie on the nine-point circle, therefore we are done. Refer to Ref. 4 for details. (P123.)
When the quadrilateral becomes cyclic, then the point of concurrency is termed as the Euler Point of the Cyclic Quadrilateral or the anticentre, in which case, the pedal circles of degenerate into the Simson lines of wrt and similars. A nice geometric proof is posted here.
Family of Properties 13.
Let be the Miquel point for the quadruple of lines , be that for the quadruple of lines and be that of Let Let and be midpoints of the segments and respectively. Similarly, define as midpoints of and be that of Let be the Newton-Gauss lines for the corresponding quadruples of lines. Then,
The lines concur at a point Similarly define
are the isogonal conjugates of with respect to triangle
are Miquel points for quadruples of lines joining
The lines are parallel.
The lines are concurrent.
The points lie on a certain circle . Similarly define circles Then, have a common point
is the point of concurrency of
AWAITING FOOL-PROOF COMPLETE PROOFS (I have not been able to solve this yet).
Let the perpendicular from the midpoint of to be denoted as Then the lines concur at a point known as the Morley point of the complete quadrilateral, and this point lies on the Miquel-Steiner line. Refer to Ref. 2 for more details.
AWAITING PROOFS WITHOUT BARYCENTRICS (Unsolved for me as well)
If the Euler line of one associated triangle is parallel to the corresponding line , then this property is true for all the four associated triangles.
Let be the centroid and orthocentre of a and let the Euler line meet the sideline at a point Let a parallel to through $ latexA$ meet at Let be the orthocentre of the Then (both perpendicular to ). Since and is also the orthocentre of the Let a parallel to through meet the line at a point The are (centrally) similar, having the corresponding sides parallel. Their similarity coefficient is -3 (because is the centroid of the and their corresponding sides are oppositely oriented). Therefore, and is the centroid of As a result, is the Euler line of the
If is any other line parallel to the Euler line of the (different from ), intersecting at the are centrally similar with center and their Euler lines are parallel, hence as well.
Let the Euler line of meet at points Let a parallel to through meet at The are centrally similar with center hence, their Euler lines are parallel. In exactly the same way as above, we can show that the Euler lines of the meet on and that
If one of the Euler lines is parallel to the corresponding line ie if then the Morley Point and the Hewer point are the same point.
We already know that the perpendiculars through the perpendicular bisectors of and similars concur at the Hewer point. And if then by PP15, we already have that these lines are the same as the lines perpendicular to the corresponding sides, and passing through the nine-point centres of the corresponding triangles, so that Morley point and Hewer points coincide for these quadrilaterals.
The polar circles of the four triangles formed by and the circumcircle of the diagonal triangle are coaxal.
Proof. (taken from Ref. 3)
Obvious from some harmonic divisions. Observe that each of the four polar circles is orthogonal to the three circles with diameter Moreover, as each of the quadruples $latex (A, U, B’,C’),(B,V,C’,A’)$ and is harmonic, the circle is orthogonal to the three circles with diameter and So this gives a shorter proof of the PP5, and also, helps us in figuring out that the orthocentric line and the pedal line are both perpendicular to the line
Family of Property 18.
We have some additional results when is a cyclic quadrilateral. We will just refer to the file “Cyclic Quadrilaterals – The Big Picture” by Yufei Zhao. [Ref. 1]
1. Cyclic Quadrilaterals – The Big Picture by Yufei Zhao.
2. Orthocentric Properties of the Plane n-Line by F. Morley.
3. Steiner’s Theorems on the Complete Quadrilateral by Jean-Pierre Ehrmann.
4. A Treatise on the Circle and Sphere by Julian Lowell Coolidge.
5. Some Properties of The Newton-Gauss Line by Catalin Barbu and Ion Patrascu.
6. Cut-The-Knot Page on Complete Quadrilaterals.
7. ENCYCLOPEDIA OF QUADRI-FIGURES by Chris Van Tienhoven (The ultimate set of properties everyone searches for).