Classical Geometry – Similarity Theorem of Triangle and Pythagoras Theorem

 

                  There are many ways to construct this system by using different postulates.  We use the following statement as Postulate ( Axiom ):

 

                                     

                       If  L1, L2, and L3 are parallel to each other,  then the line segments they intercept have the relationship:

                                                   =

 

                       If  we construct a line L6 that is parallel to L5 from the intersection of L1 and L4., we have a parallelogram.   We have a theorem in previous section that the opposite sides of a parallelogram are with equal length.

 

                      

 

                     Thus, we have the following theorem before we explore the similarity theorem of triangle:

 

Theorem:    

                            //           

 

                            

 

 

 

Definition:   Given two triangles    and   ,

                      if

                         

                             A =  D ,   B =  E ,   C =  F

                                                                       

                            

                                                                       

 

 

                   then we say     is similar to  ,  denoted as      .

 

                         

 

Theorem ( SSS ) :    If    ,  then    .

 

               Proof:  

                                        

 

1.      Without loss of generality,  we assume   >  .

On  ,  we can find a point G such that   =  .

 

2.      On point G,  we can draw a line parallel to  and we denote the intersection  of this parallel line and   as point H.

      Please see the diagram indicated above.

 

3.   //      .

       And we have   =  .  Thus,    =  and   =  .

 

4.   =  ,  =  and   = 

            (  SSS congruence theorem ).

 

5.  //        1 =  B,   2 =  C.  And  GAH =  BAC ( same one )

     plus    ,  we have     .

 

      Furthermore,     

          

 

 

Theorem ( SAS ):   If   and   A =  D,  then   .

 

                      Proof:

 

                                 

 

1.      Without loss of generality,  assume   >  .

        A =  D. Thus, it is feasible to move   to the top

        of   such that

        D is falling on A,  E is falling on  ,  and F is falling on 

        as shown in the diagram above.

 

2.          //

                              =

                           

                                                            ( SSS ).

 

 

Theorem ( AA ):  If   A =  D ,   B =  E,  then     .

 

                    Proof:

                                   

 

1.      The sum of interior angles of any triangles is 180^o ,  and   A =  D ,   B =  E

          F =    C .

 

2.  Since   A =  D,  it is feasible  to move   to the top

     of   such that

     D is falling on A,  E is falling on  ,  and F is falling on 

      as shown in the diagram above.

 

   3.   1 =  B  and   2 =   C       //

                                                        

                                                         

       Furthermore,       A =  D                ( SAS ).

 

 

 

 

 

Theorem (  Pythagoras ):    Inside ,    C = 90^o .  Then

                                            

                 ² =   ² +   ²

 

             

Proof:

                                    

1.      Draw a line from point C such that the line is perpendicular to  as

shown in the diagram above.

 

                                      2.    1 +   2 = 90^o ,   2 +   B = 90^o.

                                             Thus,    1 =  B.

                                             Furthermore,    ADC =   CDB = 90^o

 

                              ( AA )

 

                          

                         ² =    

            ( Just for reference. Will not be used for the proof )

 

          3.   Similarly,      

               (   A =  A,   ADC =  ACB = 90^o )

                              

                                 ² =     

          4.           

                  (      B =  B  ,     BDC =  BCA = 90^o  )   

                            

                                ² =  

 

          5.        ² +   ² =     +   

                                          =   (  +   )

                                          =   ²