# proof of butterfly theorem

Given that $M$ is the midpoint^{} of a chord $PQ$ of a circle and $AB$ and $CD$ are two other chords passing through $M$, we will prove that $M$ is the midpoint of $XY,$ where $X$ and $Y$ are the points where $AD$ and $BC$ cut $PQ$ respectively.

Let $O$ be the center of the circle. Since $OM$ is perpendicular^{} to $XY$ (the line from the center of the circle to the midpoint of a chord is perpendicular to the chord), to show that $XM=MY,$ we have to prove that $\mathrm{\angle}XOM=\mathrm{\angle}YOM.$
Drop perpendiculars $OK$ and $ON$ from $O$ onto $AD$ and $BC$, respectively.
Obviously, $K$ is the midpoint of $AD$ and $N$ is the midpoint of $BC$. Further,

$$\mathrm{\angle}DAB=\mathrm{\angle}DCB$$ |

and

$$\mathrm{\angle}ADC=\mathrm{\angle}ABC$$ |

as angles subtending equal arcs.
Hence triangles $ADM$ and $CBM$ are similar^{} and hence

$$\frac{AD}{AM}=\frac{BC}{CM}$$ |

or

$$\frac{AK}{KM}=\frac{CN}{NM}$$ |

In other words, in triangles $AKM$ and $CNM,$ two pairs of sides are proportional. Also the angles between the corresponding sides are equal. We infer that the triangles $AKM$ and $CNM$ are similar. Hence $\mathrm{\angle}AKM=\mathrm{\angle}CNM.$

Now we find that quadrilaterals^{} $OKXM$ and $ONYM$ both have a pair of opposite straight angles^{}.
This implies that they are both cyclic quadrilaterals^{}.

In $OKXM,$ we have $\mathrm{\angle}AKM=\mathrm{\angle}XOM$ and in $ONYM,$ we have $\mathrm{\angle}CNM=\mathrm{\angle}YOM.$ From these two, we get

$$\mathrm{\angle}XOM=\mathrm{\angle}YOM.$$ |

Therefore $M$ is the midpoint of $XY.$

Title | proof of butterfly theorem |
---|---|

Canonical name | ProofOfButterflyTheorem |

Date of creation | 2013-03-22 13:10:09 |

Last modified on | 2013-03-22 13:10:09 |

Owner | drini (3) |

Last modified by | drini (3) |

Numerical id | 6 |

Author | drini (3) |

Entry type | Proof |

Classification | msc 51-00 |