Theorem (Ptolemy): In a cyclic quadrilateral ABCD, |AB|.|CD| + |AD|.|BC| = |AC|.|BD|

What the hell you talkin' about?

Quite simple, really. If you have a cyclic quadrilateral (a four sided figure with its corners on a circle), then the sum of the products of the lengths of opposite sides equals the product of the length of the diagonals. The converse is also true. If a quadrilateral obeys the equation, then it's cyclic.

Sounds like you're making this up...

No, really. It's quite simple to prove. Watch. Say we add a point called M on the diagonal BD in such a place that the angle ACB = MCD.

                                   ,aMM"'   aP'MPVb.   `"MMa.                                   
                                 ,dM"'    ,M" ()M `Vb.    `"Mb.                                 
                               ,dM"     ,dP   ()M   `Vb.     "Mb.                 (b.           
                              aM"      aP'    M ()    `Vb.     "Ma                (PVMaa        
                            ,dP'     ,M"      M `b      `Vb.    `Vb.              ()  MP        
           ,a              ,M"     ,dP       ()  M        `Vb.    "M.             (),d"         
           (Mb            ,M'     aP'        ()  V.         `Vb.   `M.            MaP'          
           ()`b          dM'    ,M"          M   ()           `Vb.  `Mb           MMMMM)        
           d' `M.       (P    ,dP            M    M             `Vb.  V)         ,P  aP         
           M    V.     ,M    aP'            ()    M               `Vb. M.        (),d"          
           M     V)   ,M'  ,M"              ()    ()                `Vb/M.       dbP'           
          ()    dP    M' ,dP                M     `b                  `VMM       ""             
          ()  aM'    d) aP'                 M      M,aaaaaaaaaaMMMMMMMMMMMb                     
          d',d"     ,P,M"      ,aaaaaaaaaaMMMMMMMMMMP""""""""""         d'V.                    
          MaP'      MMMMMMMMMMMP"""""""""" ()        ,.                d' `b                    
         (M"       ,PM                     d'   ,.   ()               d'   V.                   
         `'        d'V.                    M    (b  ,M)              ()    `b                   
                  (P ()                   ,P    (M),P()             ,P      V)                  
                  M  `b                   ()    M VP ()            ,P        M                  
                  M   M                   d'   ,P `' M            ,P         M                  
                 ()   ()                  M    ()    M            M          ()                 
                 d'   ()                 ,P    M     M           d'          `b                 
                 M     M                 ()   (P    (P          d'            M                 
                ()     M                 d'                    d'             ()                
                ()     ()                M                    d'              ()                
                d'     `b               ,P                   ,P               `b                
                M       M               ()                  ,P                 M                
                M       V.              d'                 ,P                  M                
                M       ()              M                 ,P                   M                
                M       `b              M                 d'                   M                
                M        M             ()                d'                    M                
                M        ()            ()               d'                     M                
                M        ()            M               d'                      M                
                M         M            M              ,P                       M                
                M         V.          ()             ,P                        M                
                V.        ()          ()            ,P                        ,P                
                ()        `b          M            ,P                         ()                
                ()         M          M            M                          ()                
                 M         V.        ()           d'                          M                 
                 V.        ()        ()          d'                          ,P                 
                 ()         M        M          d'                           ()                 
                  M         M        M         ()                            M                  
                  M         ()      ()        ,P                             M                  
                  (b        ()      ()       ,P                             d)                  
                   V.        M      d'      ,P                             ,P                   
                   `b        V.     M       M                              d'                   
                    V.       ()    ,P      d'                             ,P                    
                    `b       `b    ()     d'                              d'                    
                     V)       M    d'    d'                              (P                     
                      M.      ()   M    d'                              ,M                      
                      `M.     ()  ,P   ,P                              ,M'                      
                       `M      M  ()  ,P                               M'                       
                        (b     M  d' ,P                               d)                        
                         VM.   () M ,P                              ,MP                         
                          `M.  `b,P d'                             ,M'                          
                           `Ma  M()d'                             aM'                           
                            `Vb.VdM'                            ,dP'                            
                              "MdM'                            aM"                              
                               `VMa                          aMP'                               
                                 `VMa.                    ,aMP'                                 
                            ()     `"MMa.              ,aMM"'                                   
                            dM        `"MMMbaaaaaaaaaMMM"'                                      
                           ,PM             """""""""'                                           
                           M ()                                                                 
                          () ()                                                                 
                         ,P   M                                                                 
                        M     ()                                                                
                       M'      M                                                                
We also have that the angle CAB = CDB as they lie on the same arc. Now we have two similar triangles, ABC and DMC. This means that |MD|/|DC| = |AB|/|AC| or |AB|.|CD| = |AC|.|MD| . By considering the triangles BMC and ADC in an identical way we get |AD|.|BC| = |AC|.|MB| . Adding the two equations gives us our original statement: |AB|.|CD| + |AD|.|BC| = |AC|.|MD| + |AC|.|MB| = |AC|(|MD| + |MB|) = |AC|.|BD| . Which is nice.

This has been part of the Maths for the masses project

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