Existence of Incenter, Circumcenter, Orthocenter and Center of Mass of Triangle

 

 

Theorem :  The three angle bisectors of a triangle intersect at one point.

 

           Proof:

   

                                               

 

1.      As shown on the figure on the left above,   1=  2,   3=  4.

It is sufficient to show that   5 =  6 .

 

2.      Let’s consider by using the diagram on the right above where

 

            ,    ,     .

3.  Since  O is on the bisector of    ABC, then

            

           =   .

     Similarly,     =    .

     Thus,    =  .

 

4.    OFA =  OEA = 90⁰  and   =   . 

       From Pythagorean theorem,

        ² =   ² -  ² =  ² -  ² =  ² .

 

      Thus,   =  .

5.   = ,   =  ,   = .  Then

         AOF =  AOE .

     From that, we have    5 =  6.

 

The point O is known as the “Incenter” of the triangle.

 

             Note:   Usually, we do not have  SSA  congruence theorem on triangle.  However,

                         if  the angle is 90⁰,  it can be proved that the two triangles are

                         congruent without using Pythagorean theorem . ( It is known

                         as RHS congruence theorem ).

 

 

Theorem :   The three perpendicular bisectors of sides of a triangle intersect at one point.

 

                        

 

            Proof:

1.      L and M are the perpendicular bisectors of   and  respectively.

Then  

             =  ,   =  .

So,       =   .

That implies O is on the perpendicular bisector of  .

 

2.      Thus,  the three perpendicular bisectors of three sides intersect at point O.

O is known as “Circumcenter” of  the triangle.

 

  

Lemma :    In    ABC,       =   and   =    shown as below

 

                                       

                  Then    = 2  .

 

             Proof:

1.      From point M, we draw a line which is parallel to   as shown above.

2.        //  ,   =    

           =   

3.    =   ,   = 

      So,          =  2  .

4.    //

         

 

 

Theorem:   In    ABC,       =   and   =    shown as below

                    

 

                      Then    =   .

 

                   Proof:

                                          

1.      On  , find a point Q such that   =  .

Thus,   = .

2.       = ,   =  ,   AMQ =  BMO

          AMQ =   BMO

         AQM =  BOM

     So,     //   .

3.   //  ,   =

         = 

 

The point O is known as “Center of Mass” ( Centroid ) of the triangle.

 

 

Theorem:   Three altitudes ( or their extensions ) of a triangle intersect at one point.

 

                      

                              

 

                      Proof:

                              1. As the diagram shown on the left,     ,     .

                                        And  crosses  at the point R.

                                   Then, it is sufficient to show that     to prove the theorem.

 

3.      At point C,  construct a perpendicular line to   shown

      as the diagram above.

 

    Look at   BCP and   BSC :

    it is easy to get     BCP =  BSC .

 

 

4.      Consider the points  B, C, Q, and P.

        BPC = 90⁰ =  BQC.

       Then we know that  B, C, Q, P are on one circle.

        BCP =  BQP

 

4.   APH +  AQH = 180⁰    A, P, H, Q lie on one circle .

      So,   BQP =   PAH .

 

5.      From 2, 3, 4 : we know that   PAH =  BSC.

Then   //  .

And   .

Thus,    .

 

 

The point H is known as the “Orthocenter” of the triangle.