Elementary for Classical Geometry – Sum of Interior Angles of a Triangle

 

Axiom:   If two lines L and M parallel to each other,  for any intersection N of  L and M, we have the following relationship   1 =  2 .

 

                                             

 

                  For two lines L and M are parallel to each other,  we denote as  L // M .

 

Axiom:   If   L // M // N,    for two intersections lines  shown as follows,  we have the relationship

 

                           

 

                                                                 

 

 

                   We develop the other theorems based on the two axioms.  The other theorems can be “proved” by using the two axioms here in classical geometry.  There might be some other ways to derive the whole systems. However, we just use the two axioms to construct the system.

 

                  And in ancient time, people specified the angle of a straight line as 180⁰.  It is from the convention that they divided a circle into 360  equal pieces.  We start with the following basic theorem by using this convention.

 

Theorem:  ( Opposite Angles )

                    Two lines L and M intersect at a point P as shown in the diagram.

                        

                    Then   1 =  3 .

 

Proof:              1 +  2 = 180⁰,   2 +  3 = 180⁰

                  So,    1 =  3 .

 

 

 

Theorem:  Given  ∆ABC,  prove that   A +  B +  C =  180^o

                                                                                               

 

                                

 

Proof:

 

1.      Choose a point D such that   is parallel to .  The graph is shown as above.

2.       //         1 =  C,   2 =   B .

3.       BAC +  B +  C =   3 +  2 +  1 = 180^o .

 

 

Corollary 1:   Given   shown as below:   1 =  2,  3=  4.

                      Prove that  D = 90^o +   A .

             

           Proof:

                        1.        2 +   3 +  D = 180^o .

                        2.       (  1 +  2 ) + (  3 +  4 ) +  A = 180^o .

                               And   1 =  2,   3 =  4.

                               Thus,

                                              2 +  3 = 90^o -    A .

                        3. From 1, we have  

                                  2 +   3 =    180^o -     D

 

                     90^o -    A =  180^o -     D

                      D =  90^o  +    A .

 

Corollary 2: Given    shown as below:     1 =  2,  3=  4.

                    Prove that  D =   A .

 

                  

                               

               Proof:

                             1.      (  1 +  2 ) +  BCA +  A = 180^o 　

　　　　　　　　　　Furthermore,   BCA +   3 +  4 = 180^o

                                                  　 3 +  4 = (  1 +  2 ) +  A .

 

                              2.  1 =  2,  3 =  4

                                                   2  3 = 2  2  +   A

                                                   2 =  3 -   A

 

3.      Similarly, inside

                     4 =    2 +   D

                   2 =   4 -   D

                              =   3 -  D

 

4.      From 2, 3:

                  3 -   A =     3 -  D

              D =    A

 

                          

Corollary 3:  As shown in the following diagram,    1 =  2,   3 =  4.

                    Prove that    D = 90^o -    A .

 

                 

                                                                                               

                     Proof:

1.      From , we have

          A +  ABC =   1 +  2,

          A +  ACB =   3 +  4

 

       Furthermore,   1 =  2 and  3 =  4

         2 =   (  A +  ABC )

                                                 3 =   (  A +  ACB )

 

2.      Inside  ,

                 2 +   3 +   D =  180^o

        (  A +  ABC ) +    (  A +  ACB ) +    D = 180^o

      

        A  +   (  A +  ABC +   ACB) +   D = 180^o

 

      and we have    A +  ABC +   ACB =  180^o  inside .

   

    Thus,

              A + 90^o  +   D = 180^o

 

              D =  90^o -    A