# On the Complete Quadrilateral Configurations

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 $6$ 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.

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.

Main Post.
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:

Property 1.(Miquel)
Let the lines $\ell_1,\ell_2,\ell_3,\ell_4$ 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 $M.$

Proof.
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!

$\Box$

Property 2.(Miquel Circle Theorem)
The circumcentres $O_1,O_2,O_3,O_4$ of these four triangles formed are concyclic and lie on a circle that passes through the Miquel Point.

Proof.
Refer to the diagram in Property 5. This Miquel circle doesn’t really look too hard, just note that $\angle O_3MO_2=\pi-\angle MO_2O_3-\angle MO_3O_2=\pi-\angle MDE-\angle MFE=\angle DMF.$ Using this, working backwards of PP5, it is obvious that $M,O_2,O_3,O_4$ are concyclic. So, we are done.

$\Box$

Property 3.(Miquel-Steiner Line)
The orthocentres $H_1, H_2, H_3, H_4$ of these triangles lie on a line, known as the Miquel-Steiner Line.

Proof.
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 $MH_1, MH_2, MH_3, MH_4$ lie on the pedal line mentioned in PP12, and therefore their image after homothety wrt $(M,2)$ lie on a line. This line is known as the orthocentric or Miquel-Steiner line.

$\Box$

Property 4.
The orthocentre of $O_2O_3O_4$ lies on $\ell _1.$

Proof.
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 $H_1'\in DE.$
Note that $\angle DH_1'F=\angle DH_1'O_2+\angle O_3H_1'F+\angle O_2H_1'O_3=\angle O_2O_4O_3+\angle O_2H_1'O_3=180^{\circ},$ because the reflection of $H_1'$ in $O_2O_3$ lies on the Miquel Circle.

$\Box$

Property 5.
$M$ is the centre of spiral similarity that maps $O_2O_3O_4$ to $CAB.$

Proof.
Since $O_4O_2\perp DM$ and $O_2O_3\perp ME,$ therefore $\angle O_4O_2O_3=\angle DCE$ and similarly it can be shown that $\angle O_2O_3O_4=\angle CBA.$ Therefore the triangles $O_2O_3O_4$ and $CAB$ are similar. Also, note that $\angle O_{3}MO_{4}=\angle O_{3}O_{2}O_{4}=\angle BCA=\angle BMA,$ so we see that $M$ is the centre of spiral similarity that maps $O_2O_3O_4$ to $CAB.$

$\Box$

Property 6.(Newton-Gauss Line)
The midpoints of $AC, BD, EF$ are collinear.

Proof.

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.

$\Box$

Property 7.
Define the unique point $K_1$ on $AD$ such that $BC\parallel GK_1$ and similarly define $K_2\in BC, K_3\in DC, K_4\in AB.$ Then let $M_1, M_2$ be the midpoints of $FG, DG.$ Then the points $K_i$ and $M_i$ lie on a straight line.

Proof.
Note that the sides $DA, BC$ are harmonically separated by $FG, FE.$ So, $K_2$ and $K_1$ are harmonically divided by $FG\cap K_2K_1$ and $FG\cap K_2K_1.$ Since $GK_1FK_2$ is a parallelogram, so the point $FG\cap K_2K_1$ is the point at infinity of $K_1K_1.$ Therefore $FE\parallel K_1K_2.$ So, we are done.

$\Box$

Property 8.
The projections of Miquel point onto the sidelines $AB,BC,CD,DA$ lie on a line perpendicular to the Newton-Gauss line.

Proof.
Let the points $K_i$ be defined as in our previous property. Then it can be seen that $\displaystyle \frac{\tau(K_1,\odot(FCD))}{\tau(K_1, \odot(FAB))}=\frac{\tau(K_2, \odot(FCD))}{\tau(K_2, \odot(FAB))}\Longrightarrow K_1,K_2,F,M$ are concyclic. So, $M$ is the Miquel point of both $ABK_2K_1$ and $CDK_3K_4.$
Using this, we see that projection of $M$ on $K_1K_2$ is collinear with projections of $M$ on $AB,BC,AD.$ This line is the pedal line of $ABCD.$ The fact that this is orthogonal with the Newton-Gauss line also immediately follows.

Corollary.
Using this property, it can be shown that the Miquel point lies on the Nine Point Circle of the triangle $GYZ$ where $Y=AC\cap EF$ and $Z=BD\cap EF.$ This triangle $GYZ$ is the diagonal triangle of $ABCD.$

$\Box$

Property 9.

We define the Hewer Point as the centre of the circle that passes through the orthocentres of $O_iO_jO_k$. Thus this point is the pseudo-orthocentre of $O_1O_2O_3O_4.$

Proof.
Since the point $H_P$ is defined as the circumcentre of $H_1'H_2'H_3',$ therefore it is forced that the circle $\odot(H_1'H_2'H_3')$ is congruent to the circle $\odot(O_1O_2O_3),$ and the points $H_i'$ correspond to the vertices $O_i,$ meaning that the segments $O_iH_i'$ are bisected at a common point, so that $O_1OH_1'H_P$ is a parallelogram. Therefore, $\vec{OO_1}=\vec{H_1'H_P}.$ This gives us $\vec{OH_P}=\vec{OH_1'}+\vec{OO_1} =\vec{OO_1}+\vec{OO_2}+\vec{OO_3}+\vec{OO_4},$ where the last line follows from $\vec{OH}=\vec{OA}+\vec{OB}+\vec{OC}$ in any triangle $ABC.$ So this establishes that $H_P$ is the pseudo-orthocentre of $O_1O_2O_3O_4.$

$\Box$

Property 10.
If $H_P$ is the Hewer point, $O$ the centre of the Miquel circle, $H_1'$ the orthocentre of $O_2O_3O_4;$ then we have $O_1,H_P,O,H_1'$ and $H_1', H_2', O_1, O_2$ form a parallelogram.

Proof.
Since $H_P$ is the pseudo-orthocentre of $O_1O_2O_3O_4,$ we already have that
$\displaystyle \overrightarrow{OH_P}=\overrightarrow{OO_1}+\overrightarrow{OO_2}+\overrightarrow{OO_3}+\overrightarrow{OO_4}=\overrightarrow{OO_1}+\overrightarrow{OH_1'}\Longrightarrow\overrightarrow{O_1H_P}=\overrightarrow{OH_1'};$
And all these similar properties give us our desired results.

$\Box$

Property 11.
If the Euler lines of the four triangles formed, ie $FAB, FCD, EBC, EDA$ meet the Miquel circle again at points $S_i,$ then $O_1H_P\perp MS_1$ and similars.

Proof.
Due to the spiral similarity that maps $O_1$ and $H_1'$ to $O$ and $H_1,$ note that we have $\angle$ $(MO, MO_1) =\angle$ $(OH_1'$ $, O_1H_1) ,$ and also $\vec{OH_1'}=\vec{O_1H_P},$ which leads us to the fact that $O_1H_P$ and $O_1O$ are isogonal conjugates wrt $\angle MO_1S_1.$ Because $O_1O$ passes through the circumcentre, so its isogonal passes through the orthocentre of $MO_1S_1.$ In other words, $O_1H_P\perp MS_1,$ leading to the fact that $H_P$ is the point of concurrence of the four altitudes of $MO_iS_i$ through the vertices $O_i .$

Property 12.
For an arbitrary quadrilateral $ABCD,$ the nine-point circles of triangles $BCD, CDA, DAB, ABC$ and the pedal circles of the corresponding remaining point, are concurrent.

Proof.
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.)

Special Case.
When the quadrilateral $ABCD$ 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 $A,B,C$ degenerate into the Simson lines of $A$ wrt $BCD$ and similars. A nice geometric proof is posted here.

$\Box$

Family of Properties 13.
Let $X$ be the Miquel point for the quadruple of lines $AB, AC, BD, CD.$, $Y$ be that for the quadruple of lines $AB, AD, BC, CD,$ and $Z$ be that of $BC, AC, BD, AD.$ Let $P_X=AD\cap BC, P_Y =AC\cap BD, P_Z=AB\cap CD.$ Let $K_X$ and $L_X$ be midpoints of the segments $BC$ and $AD$ respectively. Similarly, define $K_Y , L_Y$ as midpoints of $AC, BD,$ and $K_Z, L_Z$ be that of $AB, CD.$ Let $\Gamma _X=K_X L_X,$ $\Gamma _Y =K_Y L_Y,$ $\Gamma _Z=K_Z L_Z$ be the Newton-Gauss lines for the corresponding quadruples of lines. Then,
Property 13(a)
The lines $AX, BY, CZ$ concur at a point $D_0.$  Similarly define $A_0, B_0, C_0.$
Property 13(b)
$A_0, B_0, C_0, D_0$ are the isogonal conjugates of $A, B, C, D$ with respect to triangle $XYZ.$
Property 13(c)
$X, Y, Z$ are Miquel points for quadruples of lines joining $A_0, B_0, C_0, D_0.$
Property 13(d)
The lines $AA_0, BB_0, CC_0, DD_0$ are parallel.
Property 13(e)
The lines $AD, A_0D_0, YZ$ are concurrent.
Property 13(f)
The points $X,Z, P_Y, K_Y$ lie on a certain circle $\omega_Y$. Similarly define circles $\omega _X, \omega _Z.$ Then, $\omega_X, \omega_Y, \omega_Z$ have a common point $T.$
Property 13(g)
$T$ is the point of concurrency of $XP_X, YP_Y, ZP_Z.$

Proof.
AWAITING FOOL-PROOF COMPLETE PROOFS (I have not been able to solve this yet).

Property 14.
Let the perpendicular from the midpoint of $O_1H_1$ to $CD$ be denoted as $m_1.$ Then the lines $m_1, m_2, m_3, m_4$ 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.

Proof.
AWAITING PROOFS WITHOUT BARYCENTRICS (Unsolved for me as well)

Property 15.
If the Euler line of one associated triangle is parallel to the corresponding line $\ell_n$, then this property is true for all the four associated triangles.

Proof.(Yetti)
Let $G, H$ be the centroid and orthocentre of a $\triangle ABC$ and let the Euler line $GH$ meet the sideline $BC$ at a point $D.$ Let a parallel $b_1$ to $GH$ through $latexA$ meet $BC$ at $C_1.$ Let $H_1 \in AH$ be the orthocentre of the $\triangle HDB.$ Then $DH_1 \parallel CA$ (both perpendicular to $BH$). Since $(AH_1 \equiv AH) \perp (BC \equiv BC_1)$ and $BH_1 \perp (DH \parallel C_1A),$ $H_1$ is also the orthocentre of the $\triangle ABC_1.$ Let a parallel to $BC \equiv BC_1$ through $G$ meet the line $DH_1$ at a point $G_1.$ The $\triangle AC_1C \sim \triangle DGG_1$ are (centrally) similar, having the corresponding sides parallel. Their similarity coefficient is -3 (because $G$ is the centroid of the $\triangle ABC$ and their corresponding sides are oppositely oriented). Therefore, $\overline{C_1C} = -3\cdot \overline{GG_1}=3 \cdot \overline{G_1G}$ and $G_1$ is the centroid of $\triangle ABC_1.$ As a result, $(G_1H_1 \equiv DH_1) \parallel CA$ is the Euler line of the $\triangle ABC_1.$

If $b_2$ is any other line parallel to the Euler line $GH$ of the $\triangle ABC$ (different from $GH, b_1 \equiv C_1A$), intersecting $AB, BC$ at $A_2, C_2,$ the $\triangle A_2BC_2 \sim \triangle ABC_1$ are centrally similar with center $B$ and their Euler lines $G_2H_2 \parallel G_1H_1$ are parallel, hence $G_2H_2 \parallel CA$ as well.

Let the Euler line $GH$ of $\triangle ABC$ meet $CA, AB$ at points $E, F.$ Let a parallel $a'$ to $BC$ through $E$ meet $AB$ at $B'.$ The $\triangle AB'E \sim \triangle ABC$ are centrally similar with center $A,$ hence, their Euler lines $G'H' \parallel GH$ are parallel. In exactly the same way as above, we can show that the Euler lines $G'H', G_1'H_1'$ of the $\triangle AB'E, \triangle AFE$ meet on $AB'$ and that $G_1'H_1' \parallel B'E \parallel BC.$

$\Box$

Property 16.
If one of the Euler lines is parallel to the corresponding line $\ell_n,$ ie if $O_1H_1\parallel AB,$ then the Morley Point and the Hewer point are the same point.

Proof.
We already know that the perpendiculars through the perpendicular bisectors of $O_1H_1$ and similars concur at the Hewer point. And if $O_1H_1\parallel AB,$ 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.

$\Box$

Property 17.
The polar circles of the four triangles formed by $\ell_1, \ell_2, \ell_3,\ell_4$  and the circumcircle of the diagonal triangle $GYZ$ 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 $AU, BV, CW.$ Moreover, as each of the quadruples $latex (A, U, B’,C’),(B,V,C’,A’)$ and $(C, W, A',B')$ is harmonic, the circle $\odot(A'B'C')$ is orthogonal to the three circles with diameter $AU, BV$ and $CW.$ 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 $A'B'C'.$

Family of Property 18.
We have some additional results when $ABCD$ is a cyclic quadrilateral. We will just refer to the file “Cyclic Quadrilaterals – The Big Picture” by Yufei Zhao. [Ref. 1]

REFERENCES

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).

## 4 thoughts on “On the Complete Quadrilateral Configurations”

1. MarukoM says:

It is indeed true that when circles and triangles form a special bond, lines and love are born. But if we take the radius of AD and divide it K1 of M0^2, we can see that in fact some derivatives are indeed–moe. Of course we are awaiting a proof that solidifies our claim that circles and triangles can truly coexist without causing any mathematical irregularities.

2. MarukoM says:

Potla: circles and triangles can truly coexist without causing any mathematical irregularities.
MarukoM: But is still moe?
Potla: no, it won’t be planar
Potla: Cause if a graph has a K_{3,3} or K_5 in it somwhere, then it loses planarity
Potla: Some known theorem or something lol
MarukoM: That indeed fits in with my theory that tsunderes are part of z plane rather than the x or y plane
Potla: Well, they’re imaginary
Potla: They don’t fit into the normal category of people that we humans can deal with.
MarukoM: But since an imaginary tsundere is a character whose square is less than or equal to zero
Potla: You need to define seperate operations to deal with them
Potla: tsunderic multiplication
MarukoM: So one must go beyond just x y and z
MarukoM: And dwell into a new region of mathematics yet unexplored, interesting
Potla: Yeah!
Potla: There must be some t axis
Potla: They become time variant
MarukoM: I see your point. I wonder if the Gauss-Bodenmiller theorem can still apply in such a new region
Potla: That field of mathematics is kind of topology depedent
Potla: And I am yet to explore such stuff…

3. At first I didn’t understand this post, but now that I understand that this is about understanding tsunderes, who, of course, are ununderstandable, I’ve come to the understanding that it isn’t meant to be understood, at all. Well done.

4. hjklno says:

You rock!