Computer Investigation of Properties of the Gergonne Point of a Triangle
International Journal of Computer Discovered Mathematics (IJCDM)
ISSN 2367-7775 ©IJCDM
Volume 6, 2021, pp.X–X
Received XX January 2021. Published on-line XX XXX 2021
web: http://www.journal-1.eu/
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Computer Investigation of Properties

of the Gergonne Point of a Triangle

Stanley Rabinowitza and Ercole Suppab

a 545 Elm St Unit 1, Milford, New Hampshire 03055, USA

e-mail<EMAIL_ADDRESS>author

web: http://www.StanleyRabinowitz.com/

b Via B. Croce 54, 64100 Teramo, Italia

e-mail<EMAIL_ADDRESS>

Abstract. The incircle of a triangle touches the sides of the triangle in
three points. It is well-known that the lines from these points to the
opposite vertices meet at a point known as the Gergonne point of the triangle.
We use a computer to discover and catalog properties of the Gergonne point.

Keywords. triangle geometry, Gergonne point, computer-discovered mathematics,
Mathematica, GeometricExplorer.

Mathematics Subject Classification (2020). 51M04, 51-08.

##  1\. Introduction

The line from a vertex of a triangle to the contact point of the incircle with
the opposite side is called a Gergonne cevian. It is well known that the three
Gergonne cevians of a triangle meet in a point. This point is known as the
Gergonne point of the triangle. It is shown in the following illustration.

We use a computer to investigate properties of the Gergonne point. Some of the
properties found have been discovered before, and we will give a reference, if
known, to these results. In other cases, the property is new.

We systematically search for properties of the Gergonne point by starting with
a configuration that consists of a triangle and its Gergonne point. Then we
apply a sequence of 0 to 2 basic constructions to this configuration. Basic
constructions consist of typical ruler and compass constructions such as
finding the intersection of a line with a line or a circle, dropping a
perpendicular from a point to a line, drawing parallel and perpendicular
lines, constructing an incircle or circumcircle, etc. We also included
constructions involving triangle centers, such as constructing centroids,
orthocenters, and symmedian points of a triangle. Some Apollonius circle
constructions, including center and all touch points (PPP, PPL, PPC, LLL, LLP,
LLC, CLP) were also included. This includes drawing incircles, excircles,
circumcircles, and mixtilinear incircles. All possible sequences of such
constructions are generated.

Notation for Constructions
---
Notation | Description
$\triangle ABC$ | Triangle $ABC$
$a$, $b$, $c$ | The lengths of the sides of $\triangle ABC$
$s$ | $(a+b+c)/2$
$\odot ABC$ | The circle through points $A$, $B$, and $C$
$r(ABC)$ | The radius of the incircle of $\triangle ABC$
$R(ABC)$ | The radius of the circumcircle of $\triangle ABC$
$[F]$ | The area of figure $F$
$\mathtt{Gergonne}[ABC]$ | The Gergonne point of $\triangle ABC$
$\mathtt{Nagel}[ABC]$ | The Nagel point of $\triangle ABC$
$\mathtt{Spieker}[ABC]$ | The Spieker center of $\triangle ABC$
$\mathtt{orthocenter}[ABC]$ | The orthocenter of $\triangle ABC$
$\mathtt{mittenpunkt}[ABC]$ | The mittenpunkt of $\triangle ABC$
$\mathtt{Feuerbach}[ABC]$ | The Feuerbach point of $\triangle ABC$
$\mathtt{parallel}[P,AB]$ | The line through $P$ parallel to $AB$
$\mathtt{perpendicular}[P,AB]$ | The line through $P$ perpendicular to $AB$
$\mathtt{foot}[A,BC]$ | The foot of the perpendicular from $A$ to $BC$
$\mathtt{incircle}[ABC]$ | The incircle of $\triangle ABC$
$\mathtt{incenter}[ABC]$ | The incenter of $\triangle ABC$
$\mathtt{centroid}[ABC]$ | The centroid of $\triangle ABC$
$\mathtt{midpoint}[AB]$ | The midpoint of line segment $AB$
$\mathtt{ninepointcenter}[ABC]$ | The nine-point center of $\triangle ABC$
$\mathtt{colline}[A,B,C]$ | The points $A$, $B$, and $C$ colline.
$\mathtt{concur}[AB,CD,EF]$ | The lines $AB$, $CD$, and $EF$ are concurrent.

After each construction is applied, the resulting figure is analyzed for basic
properties, such as three points being collinear, three lines being
concurrent, checking for perpendicular or parallel lines, congruent incircles,
linear and quadratic relationships between lengths of line segments, etc.
Trivial properties are excluded, as well as standard properties of the
construction that is applied. The process is carried out by a program called
GeometricExplorer that we wrote specifically for this purpose. In some cases,
when GeometricExplorer found interesting numerical properties, we used a
Mathematica program to find a more general result.

Convention: Gergonne points in our figures will be colored green.

Proofs: The program we used, GeometricExplorer, works by starting with
numerical figures and examining the figures constructed using 15 decimal
places of accuracy. Although any result found is most likely true, this does
not constitute a proof. Using barycentric coordinates and Mathematica, we have
found proofs for all of the properties discovered. The proofs are in the
Mathematica notebook supplied with the supplementary material associated with
this paper.

References: A reference in blue indicates that the result was discovered by
our investigation and is believed to be new and was posted to social media in
the hopes that someone would find a purely geometric solution. An asterisk
after a reference indicates that no one has found a purely geometrical proof
as of January 2021.

##  2\. A Triangle and its Gergonne Point

In this section, we report on properties found for a starting figure
consisting of a triangle together with its Gergonne point.

###  2.1. Arbitrary Triangle

Intrinsic Properties

An intrinsic property of a figure is a property of that figure before any
constructions are applied. For example, an intrinsic property of an
equilateral triangle is that the angles all have measure $60^{\circ}$.

Property 2.1.1. (Spoke Formula) Start with: Triangle $ABC$.
$D=\mathtt{Gergonne}[ABC]$. $x=DA$. Conclusion: $\displaystyle
x=\sqrt{\frac{a(b+c-a)^{3}\left(a(a+b+c)-2(b-c)^{2}\right)}{(a^{2}+b^{2}+c^{2}-2ab-2bc-2ca)^{2}}}$.
References: See supplementary notebook.

Property 2.1.2. (Tripolar Coordinates) Start with: Triangle $ABC$.
$D=\mathtt{Gergonne}[ABC]$. $x=DA$, $y=DB$. Conclusion:
$\frac{x}{y}=\frac{b+c-a}{c+a-b}\sqrt{\frac{a(b+c-a)(a^{2}+ab-2b^{2}+ac+4bc-2c^{2})}{b(c+a-b)(b^{2}+bc-2c^{2}+ba+4ca-2a^{2})}}$.
References: This follows from the SpokeFormula. See also [12].

Property 2.1.3. (Barycentric Coordinates) Start with: Triangle $ABC$.
$D=\mathtt{Gergonne}[ABC]$. Conclusion:
$\displaystyle\frac{[BDC]}{[CDA]}=\frac{c+a-b}{b+c-a}$. References: [12]

Property 2.1.4. (Area Formula) Start with: Triangle $ABC$.
$D=\mathtt{Gergonne}[ABC]$. Conclusion:
$\displaystyle[BCD]=\frac{(a+b-c)(a-b+c)}{2ab+2bc+2ca-a^{2}-b^{2}-c^{2}}K$,
where $K=\sqrt{s(s-a)(s-b)(s-c)}$. References: See supplementary notebook.

1-step Properties

A 1-step property of a figure is a property found after applying exactly one
basic construction to the given figure.

Property 2.1.5. (Defining Property of the Gergonne Point) Start with: Triangle
$ABC$. $D=\mathtt{Gergonne}[ABC]$. Step 1: $E=\mathtt{incircle}[ABC]\cap BC$
Conclusion: $\mathtt{colline}[A,D,E]$. References: [1, p. 160], [10, p. 184]

Property 2.1.6. (LLP Property) Start with: Triangle $ABC$.
$D=\mathtt{Gergonne}[ABC]$. Step 1: A circle, center $E$, passes through $D$
and is tangent to $AB$ and $AC$. Conclusion: $DE\perp BC$. References: [25]

Property 2.1.7. (Intouch Symmedian Point) Start with: Triangle $ABC$.
$G_{e}=\mathtt{Gergonne}[ABC]$. Step 1: The incircle of $\triangle ABC$
touches $BC$ at $D$, touches $AC$ at $E$, and touches $AB$ at $F$. Conclusion:
$G_{e}$ is the symmedian point of $\triangle DEF$. References: [12]

Property 2.1.8. Start with: Triangle $ABC$. $D=\mathtt{Gergonne}[ABC]$. Step
1: The incircle of $\triangle ABC$ touches $BC$ at $E$, touches $AC$ at $F$,
and touches $AB$ at $G$. Conclusion: $R(AFD)+R(BED)=R(ABD)$. References: [48]

###  2.2. Triangles with Angle Restrictions

In this section, we report on properties found when the starting triangle has
restrictions placed on its angles.

Intrinsic Properties

No intrinsic properties were found.

1-step Properties

Property 2.2.1. Start with: Triangle $ABC$. $D=\mathtt{Gergonne}[ABC]$. Step
1: A circle, center $E$, passes through $D$ and is tangent to $AB$ at $F$ and
tangent to $BC$. Conclusion: $\angle DEF+\angle BAC=180^{\circ}$. References:
[18]

Property 2.2.2. Start with: Triangle $ABC$. $\triangle ABC$ satisfies $\angle
A=120^{\circ}$. $D=\mathtt{Gergonne}[ABC]$. Step 1: A circle, center $E$,
passes through $D$ and is tangent to $AB$ at $F$ and tangent to $BC$.
Conclusion: $\triangle DEF$ is equilateral. References: [30], special case of
previous property

Property 2.2.3. Start with: Right Triangle $ABC$. $D=\mathtt{Gergonne}[ABC]$.
$\triangle ABC$ satisfies $\angle A=75^{\circ}$ and $\angle C=15^{\circ}$.
Step 1: $H=\mathtt{orthocenter}[ACD]$. Conclusion: $[\odot ABC]+[\odot
ACD]=[\odot BDH]$. References: [29]*

Property 2.2.4. Start with: Right triangle $ACB$. $D=\mathtt{Gergonne}[ABC]$.
$\triangle ABC$ satisfies $\angle C=90^{\circ}$. Step 1: $E$ is the center of
a circle that touches $BC$ at $C$ and touches $AB$ at $G$. Conclusion:
$[BDE]=[CDG]$. References: [31]*

Property 2.2.5. Start with: Right triangle $ABC$. $\triangle ABC$ satisfies
$\angle B=90^{\circ}$. $D=\mathtt{Gergonne}[ABC]$. Step 1: The $B$-mixtilinear
incircle touches $\odot ABC$ at $E$. Conclusion: $\angle DBC=\angle DEC$.
References: [37]*

###  2.3. Triangles with Side Restrictions

In this section, we report on properties found when the starting triangle has
restrictions placed on its sides.

Intrinsic Properties

Property 2.3.1. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$a:b:c=7:9:10$. $D=\mathtt{Gergonne}[ABC]$. Conclusion: $AD=2CD$. References:
This follows from the Tripolar Coordinates.

Property 2.3.2. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$a:b:c=5:8:9$. $D=\mathtt{Gergonne}[ABC]$. Conclusion: $AD=3CD$. References:
This follows from the Tripolar Coordinates.

Property 2.3.3. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$a:b:c=4:9:10$. $D=\mathtt{Gergonne}[ABC]$. Conclusion: $AD=5CD$. References:
This follows from the Tripolar Coordinates.

Property 2.3.4. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$as=b^{2}+bc+c^{2}$. $D=\mathtt{Gergonne}[ABC]$. Conclusion: The
$A$-mixtilinear incircle of $\triangle ABC$ passes through $D$. References:
[3]

1-step Properties

Property 2.3.5. Start with: Triangle $ABC$. $\triangle ABC$ satisfies $b=2a$.
$D=\mathtt{Gergonne}[ABC]$. Step 1: The $C$-mixtilinear incircle touches $AC$
at $F$ and touches $\odot ABC$ at $E$. Conclusion: $BD\parallel EF$.
References: [32]*

Property 2.3.6. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$a=3(b-c)$. $D=\mathtt{Gergonne}[ABC]$. Step 1: $E=\mathtt{midpoint}[AC]$.
Conclusion: $[ABD]=[CED]$. References: [34]

Property 2.3.7. Start with: Right triangle $ABC$. $\triangle ABC$ satisfies
$\angle C=90^{\circ}$. $D=\mathtt{Gergonne}[ABC]$. Step 1:
$E=\mathtt{perpendicular}[D,BD]\cap BC$ Conclusion:
$\displaystyle\frac{BD}{DE}=\sqrt{\frac{2(b+c)}{c-a}}$. References: [79]

Property 2.3.8. Start with: Right triangle $ABC$. $\triangle ABC$ satisfies
$\angle C=90^{\circ}$. $D=\mathtt{Gergonne}[ABC]$. Step 1:
$E=\mathtt{perpendicular}[D,AD]\cap AC$ Conclusion:
$\displaystyle\frac{AD}{DE}=\sqrt{\frac{2bs}{(c-b)(s-a)}}$. References: [79]

Property 2.3.9. Start with: Isosceles triangle $ABC$. $\triangle ABC$
satisfies $a=c$. $D=\mathtt{Gergonne}[ABC]$. Step 1:
$E=\mathtt{parallel}[A,BC]\cap CD$. Conclusion:
$\displaystyle\frac{1}{BC}+\frac{1}{AE}=\frac{2}{AC}$. References: [74]

###  2.4. A Triangle with a Gergonne Point and Another Center

In this section, we report on properties found when including the Gergonne
point and another triangle center in the starting triangle.

Property 2.4.1. Start with: Triangle $ABC$. $\triangle ABC$ satisfies $\angle
B=60^{\circ}$. $D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{incenter}[ABC]$.
Conclusion: $\angle ABD=\angle AED$. References: [70]

Property 2.4.2. Start with: Triangle $ABC$. $D=\mathtt{Gergonne}[ABC]$.
$E=\mathtt{mittenpunkt}[ABC]$. Conclusion: $[ABDE]=[BCE]$. References: [69]

Property 2.4.3. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$2a=b+c$. $D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{Nagel}[ABC]$. Conclusion:
$\angle BAD=\angle DEG$. References: [68]

Property 2.4.4. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$2(s-a)(s-c)=b(s-b)$. $D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{Feuerbach}[ABC]$.
Conclusion: $B$, $E$, and $D$ colline and $BE=DE$. References: [87]

Property 2.4.5. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$(s-b)(s-c)=a(s-a)$. $D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{Feuerbach}[ABC]$.
Conclusion: $AE=2DE$. References: [87]

Property 2.4.6. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$a:b:c=9:7:4$. $D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{Nagel}[ABC]$.
Conclusion: $CE=DE$. References: [71]

Property 2.4.7. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$a:b:c=9:7:4$. $D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{centroid}[ABC]$.
$F=\mathtt{mittenpunkt}[ABC]$. Conclusion: $AB\perp DE$ and $AD\perp BF$.
References: [72]

Note: Our investigation found many other perpendicularities involving the
Gergonne point and another triangle center in various shape triangles. We omit
them from this report since there were so many instances.

##  3\. A Triangle with a Gergonne Cevian

In this section, we report on properties found for a starting figure
consisting of a triangle and one of its Gergonne cevians.

###  3.1. Intrinsic Properties

Property 3.1.1. (Trace Lengths) Start with: Triangle $ABC$ with Gergonne
cevian $AE$. $D=\mathtt{Gergonne}[ABC]$. Conclusion: $BE=s-b$; $CE=s-c$;
$\displaystyle\frac{BE}{CE}=\frac{s-b}{s-c}$. References: [1, p. 87]

Property 3.1.2. (Cevian Division) Start with: Triangle $ABC$ with Gergonne
cevian $AE$. $D=\mathtt{Gergonne}[ABC]$. Conclusion:
$\displaystyle\frac{AD}{DE}=\frac{a(s-a)}{(s-b)(s-c)}$. References: [1, p.
164]

Property 3.1.3. (Cevian Length) Start with: Triangle $ABC$ with Gergonne
cevian $AE$. $D=\mathtt{Gergonne}[ABC]$. Conclusion: $\displaystyle
AE^{2}=\frac{(s-a)(as-(b-c)^{2})}{a}$. References: See supplementary notebook.

Property 3.1.4. (Split Perimeter Property) Start with: Triangle $ABC$ with
Gergonne cevian $AE$. $D=\mathtt{Gergonne}[ABC]$. Conclusion: $AB+CE=AC+BE=s$.
References: This follows from the trace lengths.

Property 3.1.5. Start with: Triangle $ABC$ with Gergonne cevian $BD$.
$\triangle ABC$ satisfies $\angle B=90^{\circ}$. Conclusion: $AD\cdot
CD=[ABC]$. References: [46]

Property 3.1.6. (Kissing Circles Theorem) Start with: Triangle $ABC$ with
Gergonne cevian $AD$. Conclusion: Incircles of $\triangle ABD$ and $\triangle
ADC$ touch. References: [16]

Property 3.1.7. Start with: Triangle $ABC$ with Gergonne cevian $CD$.
$\triangle ABC$ satisfies $3a=b+c$. Conclusion: $AD=BC$. References: This
follows from the trace lengths.

Property 3.1.8. Start with: Triangle $ABC$ with Gergonne cevian $CD$.
$\triangle ABC$ satisfies $b^{2}=c(s-a)$. Conclusion: $\angle ABC=\angle ACD$.
References: [83]

###  3.2. 1-step Properties

Property 3.2.1. Start with: Triangle $ABC$ with Gergonne cevian $BF$.
$\triangle ABC$ satisfies $\angle B=45^{\circ}$ and $\angle C=30^{\circ}$.
$D=\mathtt{Gergonne}[ABC]$ Step 1: $BE$ bisects $\angle ABC$. Conclusion:
$AF=FE$. References: [27]*

Property 3.2.2. Start with: Triangle $ABC$ with Gergonne cevian $AD$. Step 1:
The $B$-mixtilinear incircle touches $\odot ABC$ at $E$. Conclusion: $AB\cdot
CE=BE\cdot CD$. References: [45]

Property 3.2.3. Start with: Triangle $ABC$ with Gergonne cevian $AD$. Step 1:
The $A$-mixtilinear incircle touches $\odot ABC$ at $E$. Conclusion:
$r(BED)=r(CDE)$. References: [23, Theorem 3.4]

Property 3.2.4. Start with: Triangle $ABC$ with Gergonne cevian $AD$. Step 1:
The incircle of $\triangle ABD$ touches $AD$ at $E$. Conclusion:
$AC+DE=AE+CD$. References: This follows from the split perimeter property.

Property 3.2.5. Start with: Triangle $ABC$ with Gergonne cevian $CD$.
$\triangle ABC$ satisfies $b+c=3a$. Step 1: $E$ is $A$-excenter of $\triangle
ABC$. Conclusion: $[AED]=[BCE]$. References: [19]

Property 3.2.6. Start with: Triangle $ABC$ with Gergonne cevian $AD$.
$\triangle ABC$ satisfies $\angle C=2\angle B$. Step 1:
$E=\mathtt{perpendicular}[D,AB].$ Conclusion:
$\displaystyle\frac{1}{BD}+\frac{1}{BE}=\frac{1}{CD}$. References: [51]

Property 3.2.7. Start with: Triangle $ABC$ with Gergonne cevian $AD$.
$\triangle ABC$ satisfies $\angle C=2\angle B$. Step 1: $E$ is the center of
the $D$-excircle of $\triangle ADC$. Conclusion: $\angle ABC=\angle AED$.
References: [52]

Property 3.2.8. Start with: Triangle $ABC$ with Gergonne cevian $AD$.
$\triangle ABC$ satisfies $\angle B:\angle C:\angle A=1:2:4$. Step 1:
$E=\mathtt{perpendicular}[AC,A]\cap BC$. Conclusion: $BD+BE=3CD$. References:
[53]

###  3.3. A Triangle with a Gergonne Cevian and Two Centers

In this section, we report on properties found for a starting figure
consisting of a triangle and one of its Gergonne cevians along with a triangle
center constructed inside each of the two triangles formed.

Property 3.3.1. Start with: Triangle $ABC$ with Gergonne cevian $AD$.
$E=\mathtt{incenter}[ABD]$. $F=\mathtt{incenter}[ADC]$. Conclusion: $AD\perp
EF$. References: [38], follows from the Kissing Circles Theorem.

Property 3.3.2. Start with: Triangle $ABC$ with Gergonne cevian $AD$.
$E=\mathtt{Gergonne}[ABD]$. $F=\mathtt{Gergonne}[ADC]$. Conclusion:
$\mathtt{concur}[AD,BE,CF]$. References: This follows from the previous
property.

Property 3.3.3. Start with: Triangle $ABC$ with Gergonne cevian $AD$.
$E=\mathtt{Nagel}[ABD]$. $F=\mathtt{Nagel}[ADC]$. Conclusion:
$\mathtt{concur}[AD,BE,CF]$. References: [39]

Property 3.3.4. Start with: Triangle $ABC$ with Gergonne cevian $AD$.
$\triangle ABC$ satisfies $b+c=ka$. $E=\mathtt{incenter}[ABD]$.
$F=\mathtt{centroid}[ADC]$. Conclusion:
$\displaystyle\frac{[BEF]}{[BCE]}=\frac{k-3}{6}$. References: [54]*

Property 3.3.5. Start with: Triangle $ABC$ with Gergonne cevian $AD$.
$\triangle ABC$ satisfies $b+c=3a$. $E=\mathtt{incenter}[ABD]$.
$F=\mathtt{centroid}[ADC]$. Conclusion: $\mathtt{colline}[B,E,F]$. References:
Special case of $k=3$ in previous property.

Property 3.3.6. Start with: Triangle $ABC$ with Gergonne cevian $AD$.
$\triangle ABC$ satisfies $b+c=3a$. $E=\mathtt{centroid}[ABD]$.
$F=\mathtt{Nagel}[ADC]$. Conclusion: Three colored regions have the same area.
References: [55]*

Property 3.3.7. Start with: Triangle $ABC$ with Gergonne cevian $AD$.
$\triangle ABC$ satisfies $b+c=3a$. $E=\mathtt{centroid}[ABD]$.
$F=\mathtt{Spieker}[ADC]$. Conclusion: $[BCE]=2[CFE]$. References: [57]

Property 3.3.8. Start with: Triangle $ABC$ with Gergonne cevian $AD$.
$\triangle ABC$ satisfies $b+c=3a$. $E=\mathtt{Gergonne}[ABD]$.
$F=\mathtt{Spieker}[ADC]$. Conclusion: $3[AEC]=8[AEF]$. References: [56]*

##  4\. A Triangle with Gergonne and Nagel Cevians

In this section, we report on properties found for a starting figure
consisting of a triangle with two Gergonne cevians or a triangle with a
Gergonne and a Nagel cevian.

###  4.1. Two Gergonne Cevians

Property 4.1.1. Start with: Triangle $ABC$. $AD$ is a Gergonne cevian. $BE$ is
a Gergonne cevian. Conclusion: $CD=CE$. References: This follows from the
Trace Properties.

Property 4.1.2. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$c=2(b-a)$. $BE$ is a Gergonne cevian. $CD$ is a Gergonne cevian. Conclusion:
$\angle CDE=\angle BAC$. References: [5]

Property 4.1.3. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$b(s-b)=2(s-c)(s-a)$. $AE$ is a Gergonne cevian. $CF$ is a Gergonne cevian.
Conclusion: $\angle ABC=\angle ADF$. References: [15]

Property 4.1.4. Start with: Right triangle $ABC$. $\triangle ABC$ satisfies
$a=3(b-c)$. $BE$ is a Gergonne cevian. $CD$ is a Gergonne cevian. Conclusion:
$\odot BCD\cong\odot ADE$. References: [81]

Property 4.1.5. Start with: Triangle $ABC$. $AD$ is a Gergonne cevian. $BE$ is
a Gergonne cevian. Conclusion: $\displaystyle\frac{[ADC]}{[BCE]}=\frac{b}{a}$.
References: [22]

Property 4.1.6. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$a:b:c=16:15:11$. $AD$ is a Gergonne cevian. $BE$ is a Gergonne cevian.
Conclusion: $AD=CE$. Note: This result is also true whenever
$4s^{3}+3abc+a^{2}b=4s(ab+bc+ca)$. References: [66]

###  4.2. A Gergonne Cevian and a Nagel Cevian

Property 4.2.1. Isotomic Conjugate Start with: Triangle $ABC$. $AD$ is a
Gergonne cevian. $AE$ is a Nagel cevian. Conclusion: $BD=CE$. References: [10,
p. 184]

Property 4.2.2. Start with: Triangle $ABC$. $AD$ is a Gergonne cevian. $AE$ is
a Nagel cevian. Conclusion: $|AB-AC|=DE$. Triangles $ABC$ and $ADE$ have the
same centroid. References: This follows from the Trace Properties.

Property 4.2.3. Start with: Triangle $ABC$. $AD$ is a Gergonne cevian. $BE$ is
a Nagel cevian. Conclusion: $AE=CD$. References: Follows from Trace Properties

Property 4.2.4. Start with: Right triangle $ABC$. $\triangle ABC$ satisfies
$\angle A=30^{\circ}$ and $\angle C=60^{\circ}$. $AD$ is a Gergonne cevian.
$CE$ is a Nagel cevian. Conclusion: $r(ADC)=r(BCE)$. References: [24]

Property 4.2.5. Start with: Triangle $ABC$. $AD$ is a Gergonne cevian. $BE$ is
a Nagel cevian. Conclusion:
$\displaystyle\frac{[ABE]}{[CED]}=\frac{2a}{b+c-a}$. References: [4]

Property 4.2.6. Start with: Triangle $ABC$. $AD$ is a Gergonne cevian. $BE$ is
a Nagel cevian. Conclusion: $\displaystyle\frac{[ADC]}{[ABE]}=\frac{b}{a}$.
References: [20]

Property 4.2.7. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$a:b:c=5:5:2$. $AD$ is a Gergonne cevian. $BE$ is a Nagel cevian. Conclusion:
$BE=2AD$ and $BN_{a}=3AG_{e}$. References: [62]

Property 4.2.8. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$a:b:c=3:2:2$. $AD$ is a Gergonne cevian. $BE$ is a Nagel cevian. Conclusion:
$BE=2AD$. Note: As in any triangle with $AB=AC$, $A$, $G_{e}$, and $N_{a}$ are
collinear. References: [63]

Property 4.2.9. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$a:b:c=9:8:3$. $AD$ is a Gergonne cevian. $BE$ is a Nagel cevian. Conclusion:
$BE=3AD$. Note: $BE=3AD$ is also true for triangles where
$a:b:c=\sqrt{41}-3:2:2$. References: [64]

Property 4.2.10. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$a:b:c=9:8:5$. $D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{Nagel}[ABC]$.
Conclusion: $AP\perp BE$ and $BQ\perp AE$. References: [65]

Property 4.2.11. Start with: Triangle $ABC$. $D=\mathtt{Gergonne}[ABC]$.
$E=\mathtt{Nagel}[ABC]$. Conclusion: $AD\perp BE$ if and only if $BD\perp AE$.
References: [86]

##  5\. A Triangle with a Cevian

In this section, we report on properties found for a starting figure that
consists of a triangle $ABC$ and one of its cevians. The cevian divides
$\triangle ABC$ into two smaller triangles. A Gergonne point is constructed in
one of these triangles or in the original triangle.

###  5.1. Gergonne Point in Original Triangle

Property 5.1.1. Start with: Right triangle $ABC$ with cevian through the nine-
point center. $\triangle ABC$ satisfies $c=2(b-a)$ and $\angle B=90^{\circ}$.
$D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{ninepointcenter}[ABC]$. Conclusion: (a)
$\mathtt{colline}[A,E,D]$. (b)
$\displaystyle\frac{AE}{ED}=\frac{3(2a+b)}{9b-14a}$. References: [77]

Property 5.1.2. Start with: Triangle $ABC$ with median. $\triangle ABC$
satisfies $c=2(b-a)$ $D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{midpoint}[AB]$.
Conclusion: $[BCD]=[CED]$. References: [2]*

Property 5.1.3. Start with: Triangle $ABC$ with median. $\triangle ABC$
satisfies $b+c=2a$. $D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{midpoint}[BC]$.
Conclusion: $(AD)^{2}+(AE)^{2}=(DE)^{2}+(BC)^{2}$. References: [33]*

Property 5.1.4. Isogonal Conjugate Start with: Triangle $ABC$ with an
$X_{55}$-cevian. $G_{e}=\mathtt{Gergonne}[ABC]$. Conclusion: $\angle
BAG_{e}=\angle CAX_{55}$ Note: $I$ is the incenter, $O$ is the circumcenter,
$M$ is the centroid, and $F_{e}$ is the Feuerbach point of $\triangle ABC$.
References: [13]

###  5.2. Gergonne Point in One of the Smaller Triangles

Property 5.2.1. Start with: Triangle $ABC$ with Nagel cevian $BD$. $\triangle
ABC$ satisfies $b+c=3a$. $E=\mathtt{Gergonne}[BCD]$. Conclusion: $CE\perp BD$.
References: [58]

Property 5.2.2. Start with: Triangle $ABC$ with symmedian $CD$. $\triangle
ABC$ satisfies $a^{2}+b^{2}=bc$. $E=\mathtt{Gergonne}[ADC]$. Conclusion:
$AE\perp CD$. References: [80]

Property 5.2.3. Start with: Triangle $ABC$ with $X_{3}$-cevian $AD$.
$\triangle ABC$ satisfies $\angle C=2\angle B$. $E=\mathtt{Gergonne}[ADC]$.
Conclusion: $CE\perp AD$. References: [59]

Property 5.2.4. Start with: Right triangle $ABC$ with angle bisector $BD$.
$\triangle ABC$ satisfies $a:b:c=3:4:5$. $E=\mathtt{Gergonne}[BCD]$.
Conclusion: (a) $\angle BEC=\angle ADB$. (b) $(DE)^{2}+(BD)^{2}=(AE)^{2}$.*
References: [40]

Property 5.2.5. Start with: Triangle $ABC$ with angle bisector $AD$.
$\triangle ABC$ satisfies $a:b:c=6:5:4$. $E=\mathtt{Gergonne}[ACD]$.
Conclusion: (a) $\angle AEC$ and $\angle ABC$ are supplementary.*
(b) $(BE)^{2}+(AC)^{2}=(BC)^{2}$.*
(c) $E=\mathtt{Nagel}[ABC]$. References: [41]

Property 5.2.6. Start with: Triangle $ABC$ with angle bisector $AD$.
$\triangle ABC$ satisfies $a:b:c=7:5:3$. $E=\mathtt{Gergonne}[ACD]$.
Conclusion: $[ABD]=[AEC]$. References: [43]

Property 5.2.7. Start with: Triangle $ABC$ with angle bisector $AD$.
$\triangle ABC$ satisfies $a:b:c=7:6:3$. $E=\mathtt{Gergonne}[ACD]$.
Conclusion: (a) $AB\parallel DE$. (b) $BD+DE=AB$. (c) $AC=9DE$. (d) $CD=7DE$.
(e) $CD+DE=2AD$. (f) $AD=4DE$. References: [42]*

Property 5.2.8. Start with: Triangle $ABC$ with angle bisector $CD$.
$\triangle ABC$ satisfies $a:b:c=9:6:5$. $E=\mathtt{Gergonne}[ACD]$.
Conclusion: $\angle ABE=\angle ACE$. References: [44]

Property 5.2.9. Start with: Triangle $ABC$ with angle bisector $AD$.
$\triangle ABC$ satisfies $a^{2}=c(b+c)$. $E=\mathtt{Gergonne}[ADC]$.
Conclusion: $DE$ is the perpendicular bisector of $AC$. References: [41]

##  6\. Lines Through the Gergonne Point

The line segment joining two points on the perimeter of a triangle is called a
chord. A chord parallel to a side of the triangle is called a parachord. This
is also known as a parallelian in the literature. The parachord through the
Gergonne point of a triangle is called a Gergonne parachord (Figure 1, left).
The line segment from the Gergonne point to the point where the Gergonne
parachord meets a side of the triangle is called a Gergonne pararadius (Figure
1, right).

###  6.1. Properties of a Pararadius

In this section, we report on properties found for a starting figure
consisting of a triangle and a Gergonne pararadius.

Figure 1. Gergonne Chord and Gergonne Pararadius

Property 6.1.1. (Pararadius Formula) Start with: Triangle $ABC$ with Gergonne
pararadius. $D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{parallel}[D,BC]\cap AC$.
Conclusion: $\displaystyle
x=\frac{-a(a+b-c)(b+c-a)}{a^{2}+b^{2}+c^{2}-2ab-2bc-2ca}$. References: See
supplementary notebook.

Property 6.1.2. (Parallel Ratio) Start with: Triangle $ABC$ with Gergonne
pararadius. $D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{parallel}[D,BC]\cap AC$.
Conclusion: $\displaystyle\frac{AE}{DE}=\frac{b}{s-c}$. References: See
supplementary notebook.

Property 6.1.3. Start with: Triangle $ABC$ with Gergonne pararadius.
$D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{parallel}[D,BC]\cap AC$. Conclusion:
$\displaystyle AE=\frac{2ab(a-b-c)}{a^{2}+b^{2}+c^{2}-2ab-2bc-2ca}$.
$\displaystyle
CE=\frac{b(b^{2}+c^{2}-a^{2}-2bc)}{a^{2}+b^{2}+c^{2}-2ab-2bc-2ca}$.
References: See supplementary notebook.

Property 6.1.4. Start with: Triangle $ABC$ with Gergonne pararadius.
$\triangle ABC$ satisfies $\angle B=90^{\circ}$. $D=\mathtt{Gergonne}[ABC]$.
$E=\mathtt{parallel}[D,BC]\cap AC$. Conclusion: $\displaystyle
BC+AE=\left(1+\frac{2b}{a}\right)DE$. References: [84]. See also [26].

Property 6.1.5. Start with: Triangle $ABC$ with Gergonne pararadius.
$\triangle ABC$ satisfies $c=3(b-a)$. $D=\mathtt{Gergonne}[ABC]$.
$E=\mathtt{parallel}[D,BC]\cap AC$. Conclusion:
$\displaystyle\frac{1}{AE}+\frac{1}{DE}=\frac{1}{CE}$. References: [78]

Property 6.1.6. Start with: Triangle $ABC$ with Gergonne pararadius.
$\triangle ABC$ satisfies $a:b:c=3:5:4$. $D=\mathtt{Gergonne}[ABC]$.
$E=\mathtt{parallel}[D,AB]\cap AC$. Conclusion: The circle through $D$
touching $AC$ at $E$ is tangent to $BC$. References: [35]

Property 6.1.7. Start with: Triangle $ABC$ with Gergonne pararadius.
$\triangle ABC$ satisfies $c=(2-\sqrt{5})a+b$. $D=\mathtt{Gergonne}[ABC]$.
$E=\mathtt{parallel}[D,AC]\cap BC$. Conclusion: $\displaystyle
x=\frac{2b(c-b)}{a-3b+c}$. References: [36] and [76]

Property 6.1.8. Start with: Triangle $ABC$ with three Gergonne pararadii.
$P=\mathtt{Gergonne}[ABC]$. $X=\mathtt{parallel}[P,AB]\cap BC$.
$Y=\mathtt{parallel}[P,BC]\cap CA$. $Z=\mathtt{parallel}[P,CA]\cap AB$.
Conclusion:
$\displaystyle\frac{1}{AZ}+\frac{1}{PX}=\frac{1}{BX}+\frac{1}{PY}=\frac{1}{CY}+\frac{1}{PZ}$.
References: [47]

###  6.2. Properties of a Parachord

In this section, we report on properties found for a starting figure
consisting of a triangle and a Gergonne parachord.

Property 6.2.1. (Parachord Formula) Start with: Triangle $ABC$ with Gergonne
parachord. $D=\mathtt{Gergonne}[ABC]$. $F=\mathtt{parallel}[D,BC]\cap AC$.
$G=\mathtt{parallel}[D,BC]\cap AB$. Conclusion: $\displaystyle
FG=\frac{2a^{2}(a-b-c)}{a^{2}+b^{2}+c^{2}-2ab-2bc-2ca}$. References: This
follows from the Pararadius Formula.

Property 6.2.2. (Parachord Perimeter) Start with: Triangle $ABC$ with Gergonne
parachord. $D=\mathtt{Gergonne}[ABC]$. $F=\mathtt{parallel}[D,BC]\cap AC$.
$G=\mathtt{parallel}[D,BC]\cap AB$. Conclusion: $AG+DF=AF+DG$. References:
This follows from the Split Perimeter Property.

###  6.3. Properties of Gergonne Chords

In this section, we report on properties found for a starting figure
consisting of a triangle and a Gergonne chord.

Property 6.3.1. Start with: Isosceles triangle $ABC$. $EF$ is a Gergonne
chord. $\triangle ABC$ satisfies $b=c$. $D=\mathtt{Gergonne}[ABC]$. $F=ED\cap
BC$. Conclusion: $m(2m+2n-p-q)(p-q)=np(p+q)$. References: [49]*

Property 6.3.2. Start with: Triangle $ABC$. $EF$ is a Gergonne chord.
$D=\mathtt{Gergonne}[ABC]$. $F=ED\cap AC$. Conclusion:
$\displaystyle\frac{s-a}{s-b}\cdot\frac{n}{m}+\frac{s-a}{s-c}\cdot\frac{q}{p}=1$.
References: [73, p. 46]

###  6.4. Perpendiculars From the Gergonne Point

In this section, we report on properties found for a starting figure
consisting of a triangle and one or more perpendiculars dropped from the
Gergonne point to the sides of the triangle.

Property 6.4.1. (Apothem Length) Start with: Triangle $ABC$.
$D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{perpendicular}[D,BC]$ Conclusion:
$\displaystyle DE=\frac{8(s-b)(s-c)K}{a(2ab+2bc+2ca-a^{2}-b^{2}-c^{2})}$,
where $K=\sqrt{s(s-a)(s-b)(s-c)}$. References: This follows from the area
formula.

Property 6.4.2. (Trilinear Coordinates) Start with: Triangle $ABC$.
$D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{perpendicular}[D,BC]$
$F=\mathtt{perpendicular}[D,AC]$ Conclusion:
$\displaystyle\frac{DE}{DF}=\frac{b(c+a-b)}{a(b+c-a)}$. References: [12]

Property 6.4.3. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$a:b:c=16:15:7$. $D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{perpendicular}[D,BC]$
Conclusion: $CD=2AE$. References: [67]

Property 6.4.4. Start with: Triangle $ABC$. $\triangle ABC$ satisfies
$a:b:c=8:5:4$. $D=\mathtt{Gergonne}[ABC]$. $E=\mathtt{perpendicular}[D,BC]$
Conclusion: $\displaystyle\frac{[BCD]}{[AED]}=632$. References: [60]

Property 6.4.5. Start with: Right triangle $ABC$. $\triangle ABC$ satisfies
$\angle A=90^{\circ}$. $D=\mathtt{Gergonne}[ABC]$.
$E=\mathtt{perpendicular}[D,AC]$ $F=\mathtt{perpendicular}[D,AB]$
$G=\mathtt{perpendicular}[D,BC]$ Conclusion:
$\displaystyle\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$. References: [61]

##  7\. Other Properties Discovered by Computer

Other properties of the Gergonne point (discovered by computer) were found by
Dekov [6]. A typical result is: The Gergonne point of a triangle is the
mittenpunkt of the orthic triangle of the intouch triangle.

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https://www.facebook.com/groups/parmenides52/permalink/3613402465440100/

  * [83] Ercole Suppa, Comment on Problem 7049. Romantics of Geometry Facebook group. Jan. 1, 2021.
https://www.facebook.com/groups/parmenides52/permalink/3613603418753338/

  * [84] Ercole Suppa, Problem 7052. Romantics of Geometry Facebook group. Dec. 31, 2020.
https://www.facebook.com/groups/parmenides52/permalink/3613742548739425/

  * [85] Ercole Suppa, Problem 7053. Romantics of Geometry Facebook group. Dec. 31, 2020.
https://www.facebook.com/groups/parmenides52/permalink/3613783122068701/

  * [86] Ercole Suppa and Stanley Rabinowitz, Problem 7066. Romantics of Geometry Facebook group. Jan. 1, 2021.
https://www.facebook.com/groups/parmenides52/permalink/3618528001594213/

  * [87] Ercole Suppa, Comment on Problem 7073. Romantics of Geometry Facebook group. Jan. 3, 2021.
https://www.facebook.com/groups/parmenides52/permalink/3619080651538948/