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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.

```

aMP'
MM"
M
,P
()
Mb.
`"VMa.
"VM)

aaaaMaaaa.
,aMMMP"VMMMP"""MMMa.
,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
MbaaaM
,P""""V)
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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