**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
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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