The Linearity Number - Oneonta such as points of concurrency for angle bisectors, medians, altitudes...

Background

Separated by approximately 1600 years, the Greek Astronomer Menelausof Alexandria (around A.D. 100) and the Italian mathematician GiovanniCeva (1647− 1736) each developed similar theorems involving collinearityof points and concurrency of lines. These discoveries turned out to betwo of the most intriguing theorems outside the mainstream of thematerial presented in Euclid’s Elements.In particular, Ceva defines the line segments joining each vertex of atriangle with a point on the opposite sides as the cevians of the giventriangle, and proves a criterion for the three cevians to be concurrent. Asit turns out, both Ceva and Menelaus work with the concept of theso-called linearity number, and they both follow the linearity numberapproach to prove the concurrency of the cevians, and the collinearity ofpoints lying on the sides (or the extended sides) of a triangle, respectively.In this presentation, we would like to talk about these two remarkabletheorems and to show how powerful they are with respect to certainapplications such as points of concurrency for angle bisectors, medians,altitudes etc, as well as the existence of the Simson’s Line.

Danielle Armaniaco The Linearity Number

Directed Distance

If l is an arbitrary line, based on the Ruler Postulate, we can alwaysconsider a coordinate system on this line so that O has the zerocoordinate.

Now, let P(x1) and Q(x2) be any two points on l , with theircorresponding coordinates x1 and x2.

P(x1) R(x

3) Q(x

2)

The directed distance from P(x1) to Q(x2) is defined as the number:

PQ = x2 − x1


Note that, in the previous definition, we no longer use the absolute valuesign. As such, we can derive the following facts:• PQ + QP = 0Indeed, (x2 − x1) + (x1 − x2) = 0• PQ + QR + RP = 0Indeed, if R(x3), then (x2 − x1) + (x3 − x2) + (x1 − x3) = 0• If PQ = PX , then Q = XIndeed, if X (x) and if q − p = x − p, then q = x , and so Q = X since,according to the Ruler Postulate, we cannot have two distinct points withthe exact same coordinates.• Finally, PQ

QR> 0 iff P − Q − R .

Indeed, if P − Q − R , then p < q < r , and so q−pr−q

> 0. Conversely, ifPQQR

> 0, then q−pr−q

> 0, which means that either both q − p > 0 andr − q > 0, case in which p < q < r , and so P − Q − R , or bothq − p < 0 and r − q < 0, which means that r < q < p, case in whichR − Q − P which is equivalent to P − Q − R .


Observation.If we construct the medians of △ABC , then they will be concurrent, and

BD = DC ,CE = EA, and AF = FB

FE

D

B

A

C

In terms of ratios, this means that

BD

DC= 1,

CE

EA= 1, and

AF

FB= 1

and so note that the product

AF

FB·BD

DC·CE

EA= 1


Question.What happens if the segments AD ,BE , and CF are not necessarilymedians? That is, what happens if, for example,

BD = 2DC and CE = 3EA,

i.e.BD

DC= 2 and

CE

EA= 3,

what should the value ofAF

FBbe so that the given segments would intersect each other at the exactsame point?

F

A

B C

D

E


The Linearity Number

The answer to the previous question is given by the Ceva’s theorem, butfirst let’s note that the product

AF

FB·BD

DC·CE

EA

will be essential to our discussions.Definition. The linearity number of the points D,E , and F with respectto △ABC is the product of the above ratios, and is denoted by thesymbol [

ABCDEF

]=

AF

FB·BD

DC·CE

EA

The points D,E , and F always lie on the sides or the extended sides of△ABC opposite to A,B, and C , respectively, and are distinct from A,B,and C .


Linearity Number Properties

First, let’s note the diagram below, which would help us easily constructthe desired ratios:

A BC

D E F

Now, let’s prove the following properties:• If we permute the columns in the symbol of the linearity number once,then we will obtain the reciprocal of the original linearity number, i.e.

[ACBDFE

]=

[ABCDEF

]−1


Indeed,

[ACBDFE

]=

AE

EC·CD

DB·BF

FA=

(CE

EA

)−1

·

(BD

DC

)−1

·

(AF

FB

)−1

which is the same as

(AF

FB·BD

DC·CE

EA

)−1

=

[ABCDEF

]−1

Even more generally, let’s note that an odd number of permutations ofthe columns will transform the linearity number into its reciprocal, whilean even number of permutations will leave it unchanged.• We also have that

[ABCDEF

]=

[ABCD ′EF

]iff D = D ′

Indeed, if D = D ′, then we obviously have

[ABCDEF

]=

[ABCD ′EF

]


Conversely, if [ABCDEF

]=

[ABCD ′EF

],

thenAF

FB·BD

DC·CE

EA=

AF

FB·BD ′

D ′C·CE

EA

which implies thatBD

DC=

BD ′

D ′C,

where D and D ′ both belong to line←→BC .

F

A

B CD=D'

E


But if we add 1 to both sides of the previous equality, we obtain that

BD + DC

DC=

BD ′ + D ′C

D ′C

which means thatBC

DC=

BC

D ′C

and soDC = D ′C .

Based on the Ruler Postulate, it will then follow that D = D ′.

We are now ready to get back to the question that we posed at thebeginning of the lecture, and, as such, to state and prove the theorem ofCeva.


The Theorem of CevaThe cevians AD ,BE , and CF of △ABC are concurrent iff

[ABCDEF

]= 1

Proof. (⇒) Suppose AD ∩ BE ∩ CF = {P}. Prove that

[ABCDEF

]= 1

We will first prove that the linearity number is positive.Indeed, if the points D,E , and F are situated on the sides of the giventriangle, then we have the following betweenness relations satisfied:

B − D − C , C − E − A, and A− F − B

F

P

B C

A

D

E


As such, based on one of the properties for directed distances,

B − D − C ⇒BD

DC> 0

C − E − A ⇒CE

EA> 0

A− F − B ⇒AF

FB> 0

and so the the product

AF

FB·BD

DC·CE

EA> 0

meaning that [ABCDEF

]> 0


Other Cases

Let’s note that if point D, for example, is no longer situated betweenpoints B and C , i.e.

B − C − D,

thenBD

DC< 0

On the other hand, let’s recall that point P is the point of concurrency ofthe given cevians. As such, we must also discuss the following threepossible situations:

A− P − D, A− D − P , and P − A− D

As before, we will prove that the linearity number is positive each andevery time (other cases would be similar).


i) Case A− P − D. In this case, ray−→CP ∈ int∠ACD, and so point

{F} =−→CP ∩

←→AB will be on the same side of line

←→AC as point D. This

means B − A− F , and soAF

FB< 0

F

E

BC

A

D

P

Also, A− P − D ⇒−→BP ∈ int∠ABC , and so point {E} =

−→BP ∩

←→AC must

be interior to side AC . This means A− E − C and, as such,

CE

EA> 0

Consequently, the linearity number[

ABCDEF

]=

AF

FB·BD

DC·CE

EA= (−) · (−) · (+) = (+)


ii) Case A− D − P . Here, ray−→BD ∈ int∠ABP , and so point

{E} =−→BP ∩

←→AC must be on the same side of line

←→BC as point P . This

means A− C − E , and therefore

CE

EA< 0

F

E

B

C

A

D

P

On the other hand, A− C − E ⇒−→PC ∈ int∠APB, and so point

{F} =−→PC ∩

←→AB must be interior to side AB . This means A− F − B and

soAF

FB> 0

As such, the linearity number[

ABCDEF

]=

AF

FB·BD

DC·CE

EA= (+) · (−) · (−) = (+)


iii) Finally, case P − A− D. This time, ray−→BA ∈ int∠PBD, and so point

{E} =−→BP ∩

←→AC will be on the same side of line

←→AB as point P , i.e.

opposite to point A as points D and C . This means E − A− C , and soCEEA

< 0.

F

E

B C

A

D

P

On the other hand, B − C − D by hypothesis, and so−→PC ∈ int∠BPD = int∠BPA. As such, point {F} =

−→PC ∩

←→AB must be

interior to side AB. This means B − F − A and therefore AFFB

> 0.We once more obtain that

[ABCDEF

]=

AF

FB·BD

DC·CE

EA= (+) · (−) · (−) = (+)


The Linearity Number Must equal 1Construct this time line

←−→B ′C ′ �

←→BC passing through point A (this can

always be done, based on the Parallel Postulate.)

E

F

D

P

B C

A B'C'

As such, we obtain the following similar triangles:

△DPB ∼ △APB ′ ⇒∣∣ BDB ′A

∣∣ =∣∣PDPA

∣∣

△DPC ∼ △APC ′ ⇒∣∣ DCAC ′

∣∣ =∣∣PDPA

∣∣

This means that

∣∣ BDB ′A

∣∣ =∣∣ DCAC ′

∣∣ or∣∣BDDC

∣∣ =∣∣B

′A

AC ′

∣∣


On the other hand,

△BCE ∼ △B ′AE ⇒∣∣CEEA

∣∣ =∣∣ BCB ′A

∣∣

△AFC ′ ∼ △BFC ⇒∣∣AFFB

∣∣ =∣∣AC

′

BC

∣∣

This means that

∣∣AFFB

·BD

DC·CE

EA

∣∣ =∣∣AFFB

∣∣ ·∣∣BDDC

∣∣ ·∣∣CEEA

∣∣ =

=∣∣AC

′

BC

∣∣ ·∣∣B

′A

AC ′

∣∣ · BCB ′A

∣∣ =∣∣AC

′

BC·B ′A

AC ′·BC

B ′A

∣∣ = 1

Consequently,∣∣[

ABCDEF

] ∣∣ = 1

and since [ABCDEF

]> 0,

it follows that [ABCDEF

]= 1


The Converse of the Theorem of Ceva

(⇐) Assume

[ABCDEF

]= 1. Prove that the cevians AD ,BE , and CF are

concurrent.

Proof. Let AD ∩ BE = {P}. Prove that the third cevian passes throughP , as well.

F=F'

P

B C

A

D

E

Indeed, let’s suppose that the third cevian−→CP intersects side AB at some

point F ′. Prove that F = F ′.


By applying the direct theorem, since AD ,BE , and CF ′ are concurrent, it

follows that

[ABCDEF ′

]= 1.

On the other hand, by hypothesis

[ABCDEF

]= 1, and so

[ABCDEF ′

]=

[ABCDEF

]

But then, based on one of the linearity number properties, this impliesthat the two points F and F ′ must coincide, and so the two cevians CFand CF ′ must coincide. As such, point P must belong to the cevian CF ,as well.


Applications: Alternate Proofs for Concurrency Points

• We already saw that the medians are concurrent, due to the simple factthat the absolute values of the ratios

∣∣BDDC

∣∣,∣∣CEEA

∣∣, and∣∣AFFB

∣∣

equal 1, and, as such, the corresponding linearity number becomes 1.

G

FE

D

B

A

C

• By looking at similar ratios we can discuss the existence of the point ofconcurrency of all of the angle bisectors, as well. As it turns out, the Lawof Sines applied to each individual angle bisector will produce very niceratios right out.


Indeed, if we apply the Law of Sines to the angle bisector AD , we obtainthat ∣∣BD

DC

∣∣ =∣∣ABAC

∣∣

I

F

D

E

B C

A

Similarly,∣∣CEEA

∣∣ =∣∣BCBA

∣∣ and∣∣AFFB

∣∣ =∣∣CACB

∣∣


Therefore

∣∣[

ABCDEF

] ∣∣ =∣∣AFFB

·BD

DC·CE

EA

∣∣ =∣∣AFFB

∣∣ ·∣∣BDDC

∣∣ ·∣∣CEEA

∣∣ =

=∣∣CACB

∣∣ ·∣∣ABAC

∣∣ · BCBA

∣∣ =∣∣CACB

·AB

AC·BC

BA

∣∣ = 1,

and since [ABCDEF

]> 0,


]= 1,

and consequently the concurrency of the angle bisectors.


Famous Points: The Gergonne Point

• The cevians joining the vertices of △ABC with the points of contact ofthe incircle on the opposite sides are concurrent. This point ofconcurrency is called the Gergonne Point of the triangle, after themathematician J.D. Gergonne (1771− 1859)

J

I'''

I'

I''

I

B C

A

In order to prove this by means of Ceva, we will start by observing thatthe center I of the incircle represents in fact the point of concurrency ofthe angle bisectors.


This is essentially due to the congruent triangles from the sketch. Startby assuming that two of the angle bisectors intersect each other at somepoint, and prove that that particular point of concurrency belongs to thethird bisector, as well. It will then follow that the lengths of theperpendiculars from the point of concurrency to the sides of the giventriangle are all the same, and, consequently, each such segment willrepresent the radius of the inscribed circle, i.e. the incircle. Moreover,the sides of the given triangle will be tangent to the incircle.

I'''

I'

I''

I

B C

A


Ceva’s Theorem Applied to Obtain Gergonne’s Point

As pointed out before, the sides are tangent to the incircle, and so thefollowing segments are congruent:

AI ′′ ∼= AI ′′′,BI ′′ ∼= BI ′, and CI ′ ∼= CI ′′

This means that

∣∣ AI′′

AI ′′′

∣∣ = 1,∣∣BI

′′

BI ′

∣∣ = 1, and∣∣ CI

′

CI ′′

∣∣ = 1

J

I'''

I'

I''

I

B C

A


Therefore

∣∣[

ABCI ′I ′′′I ′′

] ∣∣ =∣∣AI

′′

I ′′B·BI ′

I ′C·CI ′′′

I ′′′A

∣∣ =∣∣AI

′′

I ′′B

∣∣ ·∣∣BI

′

I ′C

∣∣ ·∣∣CI

′′′

I ′′′A

∣∣ = 1,

and since [ABCI ′I ′′′I ′′

]> 0,

it follows that [ABCI ′I ′′′I ′′

]= 1,

and consequently the concurrency of the stated cevians.


Concurrency of Altitudes

• As we already know, the altitudes of a given triangle are alsoconcurrent. What we may not have realized is how easily we can provethis by applying Ceva’s.

H

F

E

DB C

A

Indeed, as shown in the next sketches, we can produce three pairs ofsimilar triangles.


On one hand,

△BEC ∼ △ADC ⇒∣∣ ECDC

∣∣ =∣∣BCAC

∣∣

E

DB C

A

On the other hand,

△ABD ∼ △CBF ⇒∣∣BDBF

∣∣ =∣∣ABBC

∣∣

F

DB C

A


Finally,

△AFC ∼ △AEB ⇒∣∣AFAE

∣∣ =∣∣ACAB

∣∣

F

E

B C

A

Therefore

∣∣[

ABCDEF

] ∣∣ =∣∣AFFB

·BD

DC·CE

EA

∣∣ =∣∣AFEA

∣∣·∣∣BDFB

∣∣·∣∣CEDC

∣∣ =∣∣ACAB

∣∣·∣∣ABBC

∣∣·∣∣BCAC

∣∣

=∣∣ACAB

·AB

BC·BC

AC

∣∣ = 1

and since

[ABCDEF

]> 0, it follows that

[ABCDEF

]= 1, and consequently

the concurrency of the altitudes.


The Theorem of Menelaus

If points D,E , and F lie on the sides of △ABC , respectively, then D,E ,

and F are collinear iff

[ABCDEF

]= −1.

l

D

B

C

A

F

E

Proof. (⇒) Assume that the given points are collinear, forming line (l).

Prove that

[ABCDEF

]= −1.

We will first prove that the linearity number is negative.


Indeed, according to the Postulate of Pasch, if one of the points D,E ,and F lies on one of the sides of △ABC , then a second point must lie onanother side, while the remaining point must lie exterior to the third sideof the given triangle. For example, if A− F − B, then we must have oneof the following situations:

A− E − C and B − C − D

D

B

C

A

F

E


orB − D − C and A− C − E

E

B C

A

F

D

orB − D − C and E − A− C

E

A

BC

F

D


In either one of the previous cases, the linearity number

[ABCDEF

]=

AF

FB·BD

DC·CE

EA

will contain two positive ratios and one negative ratio. As such, it willindeed be negative.

Otherwise, all of the three points D,E , and F must lie exterior to thesides of the given triangle, as in the sketch below:

F

CB

A

D

E


In such a case,

[ABCDEF

]=

AF

FB·BD

DC·CE

EA= (−) · (−) · (−),

meaning negative once more.

Now, prove that the linearity number equals −1. For that, construct theperpendiculars from A,B, and C to produce similar triangles, as shown inthe sketch:

C'

B'

A'

D

A

B C

F

E


As such,

△AA′F ∼ △BB ′F ⇒∣∣AA

′

BB ′

∣∣ =∣∣AFFB

∣∣

△CC ′E ∼ △AA′E ⇒∣∣CEEA

∣∣ =∣∣CC

′

AA′

∣∣

and, finally,

△DCC ′ ∼ △DBB ′ ⇒∣∣DCDB

∣∣ =∣∣CC

′

BB ′

∣∣ ⇔∣∣BDDC

∣∣ =∣∣BB

′

CC ′

∣∣

This means that

∣∣[

ABCDEF

] ∣∣ =∣∣AFFB

·BD

DC·CE

EA

∣∣ =∣∣AFFB

∣∣ ·∣∣BDDC

∣∣ ·∣∣CEEA

∣∣ =

=∣∣AA

′

BB ′

∣∣ ·∣∣BB

′

CC ′

∣∣ · CC′

AA′

∣∣ =∣∣AA

′

BB ′·BB ′

CC ′·CC ′

AA′

∣∣ = 1,

and since [ABCDEF

]< 0,


]= −1


The Converse of the Theorem of Menelaus

(⇐) Assume

[ABCDEF

]= −1. Let points D,E , and F opposite to A,B,

and C , respectively. Prove that D,E , and F are collinear.

Proof. First, prove that←→DE does intersect

←→AB.

By contradiction, suppose←→DE �

←→AB. This means that we could either

have:

B − C − D ⇔ A− C − E

E

B

C

A

D


orB − D − C ⇔ A− E − C

E

B

C

A

D

orD − B − C ⇔ E − A− C

E

B C

A

D


Either way, the ratios BDDC

and CEEA

are either both positive or bothnegative. As such, the product

BD

DC·CE

EA

from the linearity number definition will be positive. On the other hand,due to the similar triangles △CDE ∼ △CBA, we have that

BD

DC=

EA

CE⇔ BD · CE = DC · EA ⇔

BD

DC·CE

EA= +1

and since, by hypothesis,

[ABCDEF

]= −1, it follows that

AF

FB= −1

which means that AF = −FB = BF . But this says that A = B, which is

impossible, so line←→DE does intersect

←→AB .


Now, let the point of intersection be some point X , as below:

l

D

B

C

A

X=F

E

This means that points D,E , and X are collinear. By applying the direct

theorem, it then follows that

[ABCDEX

]= −1. On the other hand, by

hypothesis

[ABCDEF

]= −1, and so

[ABCDEX

]=

[ABCDEF

]

But this proves that X = F , and we are done.


Applications

• As an immediate application, let us recall the centroid G of △ABC .

We can use the theorem of Menelaus to prove that G is the two-thirdspoint on BE from B to E .

G

FE

D

B

A

C

In order to prove this, all we would need to do is to extract a certain ratio

from the linearity number

[ABEGCF

]


First note that, based on Menelaus,

[ABEGCF

]= −1,

since points C ,G , and F are collinear. On the other hand,

[ABEGCF

]=

AF

FB·BG

GE·EC

CA= (+1) ·

BG

GE·(−

1

2

)

As such,BG

GE= 2 ⇔

BG

BG + GE=

2

3,

based on elementary algebra, and so

BG

BE=

2

3,

and we are done.


Another Application: The Existence of the Simson Line

The feet of the perpendiculars from any point P on the circumcircle of atriangle to the sides of the given triangle are collinear.

The line of collinearity is called the Simson line of point P for the giventriangle.

F

E

D

B C

A

P


The Pedal TriangleNow, in order to appreciate how powerful the theorem of Menelaus is inproving this result, let us first visit the classic approach to the existenceof the Simson line. We do not have to worry about directed distancesthis time, but there are other things to worry about. First, we will haveto appeal to the so called Pedal Triangle.

Definition. If we let P be an arbitrary point in the plane and if we letD,E , and F be the feet of the perpendiculars from P to the sidesBC ,AC , and AB, respectively, (extended sides, if necessary), then△DEF is called the Pedal Triangle of △ABC with respect to point P .

Observation. As P assumes various positions in the plane, the pedaltriangle will assume diverse shapes.

E

D

F

B C

A P


Formulas for the Sides of the Pedal TriangleLet BC = a, AC = b and AB = c . Also, let’s denote the distances fromP to the vertices A,B, and C , by PA = x , PB = y and PC = z .Claim.

EF =ax

2R, DF =

by

2R, and DE =

cz

2R,

where R is the circumradius of △ABC .

M

E

D

F

B C

A P


Indeed, construct the circumcircle corresponding to △AEP , and let M bethe midpoint of side AP . Note that, since

m(∠AEP) = 90◦ =1

2m(AP),

it follows that m(AP) = 180◦, so AP represents the diameter of thecircumcircle corresponding to △AEP . As such, AM represents the radius,and if we join M with E , we obtain that MA = MP = ME = r = x

2,

where r is the circumradius of △AEP .

E

M

A P


On the other hand, since m(∠AFP) = 90◦ (by hypothesis) and M is themidpoint of the hypotenuse AP , it follows once more that M isequidistant from the vertices of △AFP .

As such, MA = MP = MF = r = x2, meaning in fact that all of the

points A,E ,P , and F belong to the exact same circle of radius r2and

center M .

M

E

F

A P


Now, if we join points E and F , and extend segment FM to form FF ′, wewill obtain yet another right triangle △FEF ′, since segment FF ′ becomesthe diameter of the circle passing through A,E ,P , and F . As such,

sin(∠FF ′E ) =EF

FF ′⇔ sin(∠FF ′E ) =

EF

2r⇔ sin(∠FF ′E ) =

EF

x,

which says thatEF = x sin(∠FF ′E )

Also observe that angles ∠FF ′E and ∠FPE subtend the exact same arc,and so they are congruent.

F'

M

E

F

A

P


Get back to the original triangle △ABC . On one hand,

m(∠BAC ) +m(∠FAE ) = 180◦

On the other hand,m(∠FAE ) +m(∠FPE ) = 180◦, and so

m(∠BAC ) = m(∠FPE )

Consequently,

EF = x sin(∠FF ′E ) = x sin(∠FPE ) = x sin(∠BAC )

F'

M

E

D

F

B C

A P


One last step in obtaining the desired formulas for the sides of the pedaltriangle is to focus on the circumcircle of △ABC .

a

b

R

c

RB'

O

BC

A

Let O be the circumcenter, and R be the circumradius. Extend segmentBO to form segment BB ′. As such, BB ′ becomes the diameter of ourcircumcircle, and so △BCB ′ becomes a right triangle. Also note thatangles ∠BAC and ∠BB ′C subtend the exact same arc, and thereforethey are congruent. So

sin(∠BB ′C ) =BC

BB ′⇔ sin(∠BB ′C ) =

a

2R⇔ sin(∠BAC ) =

a

2R

⇔ EF = x sin(∠BAC ) =ax

2R


What we would like to prove next is that the pedal triangle amounts tonothing but a line when point P is situated on the circumcircle of △ABC .

F

E

D

B C

A

P

This will ultimately prove the existence of the Simson Line.


Ptolemy’s Theorem

Let us consider point P on the circumcircle of △ABC , and join P withthe vertices of the given triangle, as before. The resulting quadrilateral isa so called cyclic quadrilateral. Ptolemy’s Theorem claims that

PA · BC + PC · AB = PB · AC ⇔ ax + cz = by

a

cb

x

y

z

B C

A

P


Proof of the Ptolemy’s TheoremAssume that m(∠PCA) < m(∠ACB). Construct segment CC ′ so thatm(∠PCA) = m(∠BCC ′). This way, we produce two pairs of similartriangles. On one hand,

△PAC ∼ △C ′BC

a

c

b

x

yz

C'

B C

A

P

which will imply that

PA

BC ′=

AC

BC⇔

x

BC ′=

b

a⇔ BC ′ · b = ax


On the other hand,△PCC ′ ∼ △ACB

a

c

b

x

yz

C'

B C

A

P

which will imply that

PC

AC=

PC ′

AB⇔

z

b=

PC ′

c⇔ PC ′ · b = cz

But when we add the last two relations side by side, we obtain that

BC ′ · b + PC ′ · b = ax + cz ⇔ ax + cz = by


Back to the Simson LineWe now observe that, if we divide both sides of the previous equality by2R , we obtain that

ax

2R+

cz

2R=

by

2R,

meaning that the sides of the pedal triangle satisfy the equality

EF + DE = DF ,

which, based on the Triangle Inequality, can only occur when the pointsD,E , and F are in fact collinear.

F

E

D

B C

A

P


The Simson Line Viewed by Means of MenelausLet us get back to the linearity number

[ABCDEF

]=

AF

FB·BD

DC·CE

EA

and prove that it equals −1. The result will then follow immediately.Start by looking at △PBD and △PEA. These two triangles containangles that subtend the exact same arc. Indeed, on one hand,

∣∣ tan θ∣∣ =

∣∣PDBD

∣∣

θ

θ

F

E

D

B C

A

P


On the other hand, ∣∣ tan θ∣∣ =

∣∣PEEA

∣∣

Consequently,∣∣PDBD

∣∣ =∣∣PEEA

∣∣ ⇔∣∣BDEA

∣∣ =∣∣PDPE

∣∣

Similarly, look at △BFP and △PEC , and express tan θ′.

θ'θ'

F

E

D

B C

A

P


In the first triangle, we have that

∣∣ tan θ′∣∣ =

∣∣PFFB

∣∣,

while in the second triangle,

∣∣ tan θ′∣∣ =

∣∣PECE

∣∣

As such, ∣∣PFFB

∣∣ =∣∣PECE

∣∣ ⇔∣∣CEFB

∣∣ =∣∣PEPF

∣∣

Lastly, consider the similar triangles

△AFP ∼ △PDC ,

as shown in the following sketch.


Indeed, ∠FAP is an exterior angle for △BAP , and so

m(∠FAP) = m(∠ABP) +m(∠APB) = m(∠ACP) +m(∠ACB)

= m(∠PCD)

Also, m(∠AFP) = 90◦ = m(∠PDC ), and two pairs of congruent anglesare clearly enough to prove the required similarity.

F

D

B C

A

P

It will therefore follow that

∣∣AFDC

∣∣ =∣∣ FPPD

∣∣


We are now ready to calculate the linearity number

∣∣[

ABCDEF

] ∣∣ =∣∣AFFB

·BD

DC·CE

EA

∣∣ =∣∣AFDC

∣∣ ·∣∣BDEA

∣∣ ·∣∣CEFB

∣∣ =

=∣∣PFPD

∣∣ ·∣∣PDPE

∣∣ · PEPF

∣∣ =∣∣PFPD

·PD

PE·PE

PF

∣∣ = 1,

and since [ABCDEF

]< 0,


]= −1,

and consequently the collinearity of the feet of the perpendiculars from Pto the sides of the given triangle.


One More Famous Point: The Nagel Point• Let D,E , and F be the points of contact of the sides opposite to A,B,and C with the three excircles of △ABC . The cevians AD,BE , and CFare concurrent in a point called the Nagel Point.

N

E

F

D

A

B C


Proof

The proof is based on Ceva’s, and the fact that the sides of the giventriangle are tangent to the three excircles. As such, we can deduce quitea few equalities of lengths of tangents from each vertex, as shown below:

N

E

F

D

A

B C


We can also see that AB + x = AC + y ⇔ c + x = b + y

y

x y

x

N

E

F

D

A

B C


and since x + y = a, it follows that x = a+b−c2

, which can also be

expressed in terms of the semiperimeter s = a+b+c2

as

x = s − c ⇔ BD = s − c

As such,DC = s − b

The other tangents can be approached in a similar manner, and so

[ABCDEF

]=

AF

FB·BD

DC·CE

EA=

s − c

s − b·s − a

s − c·s − b

s − a= 1,

and consequently the concurrency follows.


The Linearity NumberDanielle Armaniaco

Faculty Advisor: Dr. Laura Munteanu

The Linearity Number - Oneonta such as points of concurrency for angle bisectors, medians, altitudes...

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Transcript of The Linearity Number - Oneonta such as points of concurrency for angle bisectors, medians, altitudes...