# Remarkable Line in Cyclic Quadrilateral: What is it about?

A Mathematical Droodle

What if applet does not run? |

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

### Explanation

In a triangle, the orthocenter H, the 9-point center N, the centroid G, and the circumcenter O are collinear. The line that houses these four and other remarkable points is known as Euler's line of the triangle. The four points always occur on the Euler line in that order: H, N, G, O. Furthermore, N is always the midway between H and O

(1) | HN:NO = 1:1 |

HG:GO = 2:1 |

What if applet does not run? |

Now, let there be a quadilateral ABCD. The four points A, B, C, and D form four triangles: ABC, BCD, CDA, and DAB, each with its own orthocenter, 9-point center, centroid, circumcenter, and Euler line. If the quadrilateral is cyclic, all four triangles share the circumcircle and the circumcenter O. In particular, in that case, the four Euler lines concur at the common circumcenter of the four triangles.

The applet illustrates the latter case of a cyclic quadrlateral. Let H_{A}, N_{A}, G_{A} denote, respectively, the orthocenter, the 9-point center, and the centroid of ΔBCD (A omitted!) Similarly introduce H_{B}, N_{B}, ..., and G_{D}. In addition to ABCD, we shall consider three more quadrilaterals: H_{A}H_{B}H_{C}H_{D}, N_{A}N_{B}N_{C}N_{D}, and G_{A}G_{B}G_{C}G_{D}.

Given the standard relative positions of the four points H, N, G, and O on the Euler line, the three quadrilaterals are not only similar, they are homothetic with the center O. H_{A}H_{B}H_{C}H_{D} is obtained from G_{A}G_{B}G_{C}G_{D} with a factor of 3, while N_{A}N_{B}N_{C}N_{D} is obtained from G_{A}G_{B}G_{C}G_{D} with a factor of 3/2.

In any quadrilateral ABCD, the quadrilateral of centroids G_{A}G_{B}G_{C}G_{D} is homothetic to ABCD with a factor of -1/3. This is an interesting exercise to show that the product (successive application) of two homotheties is either a homothety or a translation. In the former case, the factor of the product is the product of the factors of the constituent homotheties. Furthermore, the three centers of homothety are collinear.

As an application of this result, we find that H_{A}H_{B}H_{C}H_{D} is the homothetic image of ABCD with the factor of _{A}N_{B}N_{C}N_{D} is the homothetic image of ABCD with the factor of

Let H, N, and G be the centers of homothety of ABCD and H_{A}H_{B}H_{C}H_{D}, N_{A}N_{B}N_{C}N_{D}, and G_{A}G_{B}G_{C}G_{D}, respectively. From the above remark, the four points H, N, G, and O are collinear. In itself, this is a curious property of a cyclic quadrilateral. We may go a little further and determine the relative locations of the points H, N, G, and O on that line.

Perhaps surprisingly the ratios are different from (1):

(2) | HN:NO = 1:2 |

HG:GO = 1:1 |

I'll give two proofs of that fact. One is based on a theory of homothetic transformations, the other that was communicated by Paul Yiu makes use of complex numbers.

Let two homotheties -- the first with center S_{1} and factor k_{1} and the second with center S_{2} and factor k_{2} -- are carried out successively. Barring exceptional cases, the result is a homothety with factor k_{1}k_{2}. The question is to determine the center S of that homothety. As we know, the three points S_{1}, S_{2}, and S are collinear.

Let S' be the image of S_{1} under the composite homothety. This means that S'S = k_{1}k_{2}·S_{1}S. On the other hand, the first homothety leaves S_{1} fixed, while the second expands S_{1}S_{2} by the factor of k_{2}, such that S'S_{2} = k_{2}S_{1}S_{2}. Subtraction gives

S'S - S'S_{2} = k_{1}k_{2}·S_{1}S - k_{2}S_{1}S_{2},

or

S_{2}S = k_{1}k_{2}·(S_{1}S_{2} + S_{2}S) - k_{2}S_{1}S_{2},

and, finally,

S_{2}S = (k_{1} - 1)k_{2}/(1 - k_{1}k_{2})·S_{1}S_{2}.

In a somewhat simpler form we have

(3) |
S_{1}S = (1 - k_{2})/(1 - k_{1}k_{2})·S_{1}S_{2}. |

Let's apply (3) in our case:

S_{1} | k_{1} | S_{2} | k_{2} | S | (3) | |||||
---|---|---|---|---|---|---|---|---|---|---|

G | -1/3 | O | 3 | H | GH:GO = -1:1 | |||||

G | -1/3 | O | 3/2 | N | GN:GO = -1:3 |

which is equivalent to (2).

### References

- I. M. Yaglom,
*Geometric Transformations II*, MAA, 1968

For the second approach, let's put the circumcenter O at the origin, and represent the vertices A, B, C, D of the quadrilateral by unit complex numbers a, b, c, d.

It is well known that the orthocenter of triangle xyz is x + y + z etc.

We consistently denote by P(t) the point P on the Euler line of a triangle with

P(t) of triangle bcd is t(b + c + d), | |

P(t) of triangle cda is t(c + d + a), | |

P(t) of triangle dab is t(d + a + b), and | |

P(t) of triangle abc is t(a + b + c). |

The quadrilateral of P(t)'s of the four triangles bcd, cda, dab, abc is homothetic to the quadrilateral abcd with factor -t. The center of homothety is the point Q(t) such that

Q(t) = t(a + b + c + d)/(1 + t).

Note that (a + b + c + d)/4 is the centroid G of the quadrilateral, so that this center of homothety traverses the line OG as claimed. Furthermore,

(4) | H = Q(1) = (a + b + c + d)/2, |

N = Q(1/2) = (a + b + c + d)/3, and, finally, | |

G = Q(1/3) = (a + b + c + d)/4, |

which implies (2).

### Remark

From (4) it follows that Q(1) -- the center of homethety of ABCD and H_{A}H_{B}H_{C}H_{D} -- serves as the circumcenter of N_{A}N_{B}N_{C}N_{D}. Indeed, N_{A} is represented by the complex number (b + c + d)/2, so that

We see that the 4 nine point circles meet in H. We also know that the four simsons constructed for the four triangles ABC, BCD, CDA, and DAB with the points D, A, B, and C, respectively, concur with the nine point circles, i.e., at H.

Darij Grinberg [see, The Euler point of a cyclic quadrilateral] has also observed that the maltitudes of ABCD concur in the same point H, which makes it the anticenter of the quadrilateral. To see why this is so, let M_{XY} denote generically the midpoint of segment XY. Then

BH_{A} ⊥ CD, as an altitude in ΔBCD,

H = M_{AHA}, for H is the center of homothety with k = 1,

M_{AB}H || BH_{A}, as a midline in ΔABH_{A}.

Thus M_{AB}H ⊥ CD, and the same holds for other maltitudes.

Now observe that the same four points A, B, C, D define quadrilaterals ABCD, ACBD, and ACDB. The foregoing discussion concerning ABCD applies as well to the latter two. In particular, the maltitudes from AC and BD concur with the maltitudes from AD and BC. Turning our attention back to ABCD, this means that the perpendicular from the midpoint of a diagonal to the other diagonal is also incident to H.

A related result has been established for more general quadrilaterals. Some other four points H', N', G', and O' happen to be collinear, lie in the same order, and obey the same proportions.

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny