# proof of Simson’s line

Given a $\mathrm{\u25b3}ABC$ with a point $P$ on its circumcircle^{} (other than $A,B,C$),
we will prove that the feet of the perpendiculars^{} drawn from P to the sides $AB,BC,CA$
(or their prolongations) are collinear^{}.

Since $PW$ is perpendicular to $BW$ and $PU$ is perpendicular to $BU$ the point $P$ lies on the circumcircle of $\mathrm{\u25b3}BUW$.

By similar^{} arguments, $P$ also lies on the circumcircle of $\mathrm{\u25b3}AWV$ and $\mathrm{\u25b3}CUV$.

This implies that $PUBW$ , $PUCV$ and $PVWA$ are all cyclic quadrilaterals^{}.

Since $PUBW$ is a cyclic quadrilateral,

$$\mathrm{\angle}UPW={180}^{\circ}-\mathrm{\angle}UBW$$ |

implies

$$\mathrm{\angle}UPW={180}^{\circ}-\mathrm{\angle}CBA$$ |

Also $CPAB$ is a cyclic quadrilateral, therefore

$$\mathrm{\angle}CPA={180}^{\circ}-\mathrm{\angle}CBA$$ |

(opposite angles in a cyclic quarilateral are supplementary^{}).

From these two, we get

$$\mathrm{\angle}UPW=\mathrm{\angle}CPA$$ |

Subracting $\mathrm{\angle}CPW$, we have

$$\mathrm{\angle}UPC=\mathrm{\angle}WPA$$ |

Now, since $PVWA$ is a cyclic quadrilateral,

$$\mathrm{\angle}WPA=\mathrm{\angle}WVA$$ |

also, since $UPVC$ is a cyclic quadrilateral,

$$\mathrm{\angle}UPC=\mathrm{\angle}UVC$$ |

Combining these two results with the previous one, we have

$$\mathrm{\angle}WVA=\mathrm{\angle}UVC$$ |

This implies that the points $U,V,W$ are collinear.

Title | proof of Simson’s line |
---|---|

Canonical name | ProofOfSimsonsLine |

Date of creation | 2013-03-22 13:08:26 |

Last modified on | 2013-03-22 13:08:26 |

Owner | giri (919) |

Last modified by | giri (919) |

Numerical id | 9 |

Author | giri (919) |

Entry type | Proof |

Classification | msc 51-00 |