Classical Geometry – Congruence Theorem of Triangle

 

Definition:  For   and    ,

                  if 

 A =  D,    B =  E ,   C =  F

 

 =  ,   =   ,   = 

                 then we say   is congruent to ,  denoted as     .

                                                                       

 

 

                      

 

 

Theorem ( SSS ):    If   =  ,   =   ,   =  ,  then      .

      

             Proof: 

 

                                

                                            

 

                              1.    =   .  Thus, it is feasible to  move    to the place

                                     such that   and   are overlapped together with

                                     point A falling on point D and point B falling on point E.

 

2.      Use the length of   as radius and point B as center to draw a circle.  Point C must be on the circle

because   =   .

 

3.      Similarly, use the length of   as radius and point A as center to draw a circle. 

 

      Assume that Point C is not on this circle.   Let’s say that the

      circle crosses  at the point G.  G  C .

 

4.      If  point G is between  point A and C,  then   >   .  But  we know that   =  .

Thus, this case does not hold.

 

If point G is not between A and C, but G is on   ,

 then   <   .  It contradicts with the fact that   =  .   This case doe not hold either.

 

So, the point G must overlap with point C.  In the way,   and  can be overlapped together  with D falling on A, E falling on B, and F falling on C.   Thus, we get the

conclusion that      .

                                                                       

 

 

Theorem ( SAS ) :      If   =  ,   A =  D ,   =  , 

                                    then     .

 

           Proof:

 

                                   

 

                               1.   A =  D.   Thus, it can be done by moving   to

                                     the top of    with   being tied with   , 

                                     and   being tied with   ,  shown as the diagram above.

                                     Otherwise,    A =  D does not hold.

 

2.      Use point B as the center and the length of   to draw a circle.  If the circle has no intersection with   ,  it means that using the lengths  ,    ,  and   can not form a triangle.   Thus, the circle must intersect with  .   Let’s call the intersection point G.

Assume  G   C .

 

3.      If  G is between  point A and C,  then  >  . ( contradiction )

If G is not between A and C but G is on  ,  then  <  . ( contradiction )

 

Thus, G must fall on C.  In other words,   =  .

 

4.      With  =  , we also have   =   and   =  .

Then     ( via SSS ).

 

                                                                       

 

Theorem ( ASA ) :   If  B =  E ,    =  ,  C =  F,  then    .

 

                Proof:

 

                                

 

                              1.    B =  E ,    =  ,  C =  F.  Thus, it is feasible

                                     to overlay   on  with E falling on B, 

                                     F falling on C,  and  D falling on  .   If  D does not fall

                                      on  ,  the statement   B =  E can not be true.

                                      Assume D  A .

 

2.      If  D is between point B and A,  then   BCA >   EFD.  ( contradiction )

If D is not between point B and A, but D is on ,

then   BCA <   EFD.

That also contradicts to the fact that  C =  F.

 

Thus,  D = A.      =  .

 

                                3.   =  ,   B =  E ,    =    

                                           ( via SAS ).

 

 

Theorem ( AAS ):   If   A =  D,    B =  E, and   =  ,

                                then    .

 

        Proof:

                It is trivial by using the fact that  the sum of interior angles of a triangle is 180^o.