Classical Geometry – Circle II

 

Theorem:    In the following diagram,     = 

                                                                       

 

                                 

                   Proof:

                                        

                                    1.  Construct   and  as above.

                                    2.    CAB =  CDB

                                      (    Both of them are the inscribed angles of   ).

                                           AEC =  DEB

                                          Thus,       ( AA ).

                                          This implies

                                                                  

                                          So,   = 

 

 

                      Note:     CEB =   ABD +   CDB (  check  )

                                   Think about the relationship with the arcs…

 

 

Theorem:               As shown in the following diagram, prove       =    .

 

                          

                    Proof:

                                               

1.      Connect  ,  .

2.     A =  C  ( corresponding to the same arc  ),  and   E =  E

         

 

         Thus,  

                       

         So,      =    .

              

                     Note:    E =   ADC -  A

 

 

Theorem:          A, B, C, D are on the same circle as the diagram below.

                       Then   ABC =  CDF .

                

                   Proof:

                                  1.    ABC +  ADC =  180^o  (  Their arcs cover the whole circle ).

                                  2.   ADC +   CDF = 180^o  ( A, D, F are on the same line )

                                  3. From 1, 2: we have   ABC =   CDF .

 

 

 

Definition:  ( Tangent  Line of a circle)

                  If a line crosses a circle at one and only one point, then this line is a tangent line of the circle.

 

 

 

 

 

 

 

Theorem:   As the diagram shown,  O is the center of the circle and L is a tangent line of the circle by crossing the circle at point A.   Then     L .

                                                                       

 

                   

                 Proof:

 

                                                 

1.      Assume   is not perpendicular to the line L.

Then we can dram a line from O such that this line is perpendicular to L and

crossing L at point H.   H  A.

2.      Consider  .   > .

But   is the radius of the circle.  Thus, there must exist the other point such that

the line L crosses the circle at that point.

However, L only crosses the circle at point A. (  contradiction )

 

So, the assumption does not hold.  Then  is perpendicular to the line L.

 

 

Theorem:    A, B, C are on the circle.  L is a tangent line that crosses the circle at point A as below.

                   Then   BAD =  BCA .

                                       

                   Proof:

                                        

1.      As shown above,  O is the center of the circle and   is the diameter.

Then     .    AEB +  EAB = 90^o .

                                  2.  L is a tangent line.  So, we have     L .

                                          EAB +  BAD = 90^o .

                                      

                                        So,    BAD =   AEB =  ACB .

                                                The last equality holds because they extend from the same arc.

 

 

 

Theorem ( Ptolemy ):

 

                                As shown in the following diagram, prove that

                                          +      =                                   

                                                                       

                              

                      Proof:

                                         

1.      From point C, we can draw a line that intersects with  at point F such that

                                            DCF =   BCA ,  as shown in the diagram above.

2.      A, B, C, D are on the same circle.   So, we have

        CDF =  CBA.

3.    DCF =   BCA ,   CDF =  CBA

             ( AA )

         

        So,     =

 

4. Similarly,    BCD =  BCA +  ACD =   DCF +  ACD =  ACF

5.    CAD =   CBD  ( corresponding to the same arc )

6.    CAD =   CBD,   BCD =  ACF

          

       

      Thus,   =   =   (  +   )

 

7. Summarizing the result from 3 and 6, we have

          =   +   =   +