The  Proof  of  Girard's  Theorem    Mathematics:  Exploration  Fidan  Mammadli  

Session  number:  004423-­‐0008  

 

   

  2  

Girard's theorem

Introduction:

My interest in non-Euclidian geometry, especially in Spherical geometry affected my

choice of topic. Therefore, in this exploration I would like to focus on proof of Girard's

theorem. However, firstly I would like to define necessary key terms and derive some

other formulas of Spherical geometry. This work will consist of three steps. In the first

step I will get expression for the surface area of spherical lune and in the second step I

am going to prove the equality of antipodal triangles. The conclusions from these two

steps will be used in the third step where Girard's theorem will be finally proved. In order

to make the explanation clear I will include some hand-drawn sketches. However, I also

decided to use SketchUp Make 2015 software because for the exploration it is important

to show some 3D pictures from different points of view, which would be quite difficult to

draw on paper.

  3  

Key  terms:  

• Great circle - A circle drawn on a sphere with centre coinciding with centre of sphere

and which has radius equal to that of sphere. (Figure 1)

• Antipodal points - Two diametrically opposite points on a sphere as points A and B in

Figure 2.

• Lune - Area on the surface of sphere enclosed by two semicircles intersecting at

antipodal points on a sphere. (Figure 3)

                                                                                                                   Figure  1                                                                                                                                      Figure  2  

 

                                                                                                                                                                                                                         Figure  3                                                                                                              

                                                                   

  4  

Step  №1:  

Aim:  to  prove  that  area  of  Lune  is  2θ𝑅!  

Let's consider a particular lune created by 2 great circles which will enclose 1 4 the

sphere, which  therefore  has  got  angles  equal  to  !!!=!!.  (Figure  1.1)

Figure 1.1 If  we  denote  Area  of  Sphere  by  𝑨  and  the  Area  of  this  lune  by  S   then  S  will  be  

equal  to  𝑨!:  

A= 4𝜋𝑅!  è S=!!=    !!!!  !

=2  !!𝑅2  

To  derive  the  formula  for  the  area  of  any  given  lune  we  should  denote  its  angles  

by  𝜽.  So   therefore   in   this  particular  case  𝜽 = 𝜋2.   By  substituing  

!!  with  𝜽  in   the  

formula  above  we  will  get:  

S=𝟐𝜽𝑹𝟐  

Conclusion  1:  Area  of  any  lune  created  by  semicircles  of  2  great  circles  can  be  

calculated  by  the  formula  S=𝟐𝜽𝑹𝟐,  where  R  is  radius  of  sphere  and  is  𝜽  angle  of  

the  lune.  

  5  

Step  №2:  Aim:  to  prove  that  areas  of  antipodal  triangles  are  equal.  

Antipodal  triangles  are  those  triangles  that  are  formed  by  intersections  of  same  

3  great  circles.  Each  vertex  of  one  triangle  has  got  its  antipodal  point,  which  is  

vertex  of  another  one.  For  example,  yellow  triangle  △ABC  and  blue  △A'B'C'   in  

Figure  2.1.  

     

Figure  2.1  (The  figure  is  3D,  dashed  lines  demonstrate  image  on  opposite  side  of  the  sphere)  

 

  6  

Each  side  of   these  two  triangles   is  an  arc  of  one  of   the  3  great  circles.  So  

lets  look  at  3  different  cross  sections  of  the  sphere,  which  are  hemispheres  LM,  

NS  and  KP.  

1. Consider   hemisphere   LM   in   Figure   2.2.   The   image   drawn   at   Right  

Hand  Side  shows  viewpoint  from  which  the  figure  is  displayed.  

 Figure  2.2  

Length of arc= r𝜔 (where 𝜔 is measure of central angle and r is raduis of the circle)

In this case our central angles are ∠𝐵𝑂𝐶  and  ∠𝐵′𝑂  𝐶′. As they were created by

intersection of two diameters and they are opposite angles, it is obvious that ∠𝐵𝑂𝐶

and ∠𝐵′𝑂  𝐶′ are equal.

∠𝐵𝑂𝐶 = ∠𝐵′𝑂  𝐶′ = 𝛼 è BC=  R𝛼  and  B'C'=  R𝛼      è BC=B'C'  

 

 

  7  

2. By  the  same  method  we  can  consider  hemisphere  KP.  (Figure  2.3).  

 

Figure 2.3

Angles ∠𝐵𝑂𝐴  and  ∠𝐵′𝑂  𝐴′ are central and opposite angles, therefore are equal.

∠𝐵𝑂𝐴 = ∠𝐵′𝑂  𝐴′ = 𝛽   è BA=  R𝛽  and  B'A'=  R𝛽      è BA=B'A'  

                   

  8  

 3.  Let's  repeat  the  procedure  in  previous  two  cases  and  look  at  the  3rd  

hemisphere  SN.  (Figure  2.4)  

 

 Figure 2.4

∠𝐶𝑂𝐴 = ∠𝐶′𝑂  𝐴′ = 𝛾è CA=  R𝛾  and  C'A'=  R𝛾      è CA=C'A'      

 

Conclusion  2:  AB=A'B';  BC=B'C';  CA=C'A'  è Each  side  of  yellow  triangle  

△ABC  is  equal  to  one  side  of  blue  triangle  △A'B'C'  which  means  that  these  

 triangles  are  equal  and  have  same  areas.    

△ABC=△A'B'C'        

     

  9  

Step  №3:  

 Aim:  to  derive  general  formula  for  the  Area  of  spherical  triangle  and  prove  

Girard's  theorem.  

Girard's   theorem:  A spherical triangle, drawn on the surface of a sphere of

radius R and having angles 𝜶,𝜷,𝜸 has area A  =(𝜶+𝜷+𝜸 −  𝝅)  𝑹𝟐.  

I  used  Sketch  Up  Make  2015  software  to  

display   from   different   points   of   view   a  

sphere  with  3  great  circles,  which  create  

2  antipodal  triangles  and  6  lunes.  (Black  

and   White   areas   are   our   antipodal  

triangles   and   Areas   of   the   same   colors  

are  equal  because  they  are  antipodal  duplicates)  

  10  

It is important to understand that each triangle is a part of 3 different lunes. The

White triangle creates lunes with Purple, Yellow and Green areas whereas Black

triangle creates lunes with their antipodal duplicates. Therefore each angle of the

triangles are equal to the angle of one of the lunes:

Lune 1 Lune 2

Lune 3 Lune 4

Lune 5 Lune 6

  11  

Let's denote the area of white trinagle by T and the area of black triangle by T'.

From Conclusion 2 we know that these triangles are equal because they are antipodal

duplicates of each other, therefore:

T = T'

• We will denote angles of triangles by 𝜶𝟏;𝜶𝟐;𝜶𝟑.

• Let's denote Lunes by 𝐿! where i is the number of the lune and i∈ 1; 6 .

Taking into account that areas of the same colors are also equal it can be concluded

that:

L1= L4

L2= L5

L3= L6

• If all these Lunes are added up it will count areas of each triangle 3 times,

whereas to get surface area of sphere we need each triangle only once, so:

𝐿!  =!

!!!

 𝐀𝐫𝐞𝐚  𝐨𝐟  𝐬𝐩𝐡𝐞𝐫𝐞   +  𝟐𝐓   +  𝟐𝐓′

• From Conclusion 1 we know that area of the lune is equal to 2θ𝑅!. An

angle of each lune is equal to one angle of triangle. Therefore:

𝑳𝟏 = 𝑳𝟒 =  2𝛼!𝑅!  

𝑳𝟐 = 𝑳𝟓 =  2𝛼!𝑅!  

𝑳𝟑 = 𝑳𝟔 =  2𝛼!𝑅!

  12  

𝑳𝒊  =𝟔𝒊!𝟏 𝟒𝜶𝒊𝑹𝟐 =𝟑

𝒊!𝟏  4𝜶𝟏𝑹𝟐+4𝜶𝟐𝑹𝟐+4𝜶𝟑𝑹𝟐                

4𝜶𝟏𝑹𝟐+4𝜶𝟐𝑹𝟐+4𝜶𝟑𝑹𝟐    = Area of sphere + 2T + 2T'      

4𝜶𝟏𝑹𝟐+4𝜶𝟐𝑹𝟐+4𝜶𝟑𝑹𝟐    = 𝟒𝝅𝑹𝟐+ 4T  

4T  =4𝜶𝟏𝑹𝟐+4𝜶𝟐𝑹𝟐+4𝜶𝟑𝑹𝟐 −  𝟒𝝅𝑹𝟐  

T  =4𝜶𝟏𝑹𝟐+4𝜶𝟐𝑹𝟐+4𝜶𝟑𝑹𝟐 −  𝝅𝑹𝟐  

T=(𝜶𝟏 + 𝜶𝟐 + 𝜶𝟑 −  𝝅)  𝑹𝟐          

                 

Conclusion:  

The calculations above gave me formula of Girard's theorem, which I was expecting

to get.

Therefore, Girard's theorem which stated that a spherical triangle, drawn on a sphere

of radius R and with angles 𝜶𝟏,𝜶𝟐,𝜶𝟑 has an area A =(𝜶𝟏 + 𝜶𝟐 + 𝜶𝟑 −  𝝅)  𝑹𝟐 is

proven to be correct. This   formula   is   obvious   evidence   of   that   spherical  

geometry  significantly  differs  from  classic  Euclidian  geometry.  For  example,  the  

rule  that  interior  angles  of  polygon  should  be  equal  to  180°  is  not  applicable  for  

spherical  geometry.  That  is  why  the  fact  that  formula  for  Spherical  triangle  is  so  

different  from  Euclidian  is  logical.    

 References:  

1. http://www.theoremoftheday.org/GeometryAndTrigonometry/Girard/TotDG

irard.pdf

2. http://www.princeton.edu/~rvdb/WebGL/GirardThmProof.html

3. http://math.rice.edu/~pcmi/sphere/gos4.html