# Heron’s principle

Theorem. In the Euclidean plane^{}, let $l$ be a line
and $A$ and $B$ two points not on $l$. If $X$ is a point of
$l$ such that the sum $AX+XB$ is the least possible,
then the lines $AX$ and $BX$ form equal angles with the
line $l$.

This Heron’s principle, concerning the reflection^{} of
light, is a special case of Fermat’s principle in optics.

Proof. If $A$ and $B$ are on different sides of $l$, then $X$ must be on the line $AB$, and the assertion is trivial since the vertical angles^{} are equal. Thus, let the points $A$ and $B$ be on the same side of $l$. Denote by $P$ and $Q$ the points of the line $l$ where the normals of $l$ set through $A$ and $B$ intersect $l$, respectively. Let $C$ be the intersection point of the lines $AQ$ and $BP$. Then, $X$ is the point of $l$ where the normal line^{} of $l$ set through $C$ intersects $l$.

Justification: From two pairs of similar^{} right triangles^{} we get the proportion equations

$$AP:CX=PQ:XQ,BQ:CX=PQ:PX,$$ |

which imply the equation

$$AP:PX=BQ:XQ.$$ |

From this we can infer that also

$$\mathrm{\Delta}AXP\sim \mathrm{\Delta}BXQ.$$ |

Thus the corresponding angles $AXP$ and $BXQ$ are equal.

We still state that the route $AXB$ is the shortest. If ${X}_{1}$ is another point of the line $l$, then $A{X}_{1}={A}^{\prime}{X}_{1}$, and thus we obtain

$$A{X}_{1}B={A}^{\prime}{X}_{1}B={A}^{\prime}{X}_{1}+{X}_{1}B\geqq {A}^{\prime}B={A}^{\prime}XB=AXB.$$ |

## References

- 1 Tero Harju: Geometria. Lyhyt kurssi. Matematiikan laitos. Turun yliopisto (University of Turku), Turku (2007).

Title | Heron’s principle |
---|---|

Canonical name | HeronsPrinciple |

Date of creation | 2014-09-15 15:38:36 |

Last modified on | 2014-09-15 15:38:36 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 13 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 51M04 |

Related topic | Catacaustic |

Related topic | PropertiesOfEllipse |

Related topic | HeronianMeanIsBetweenGeometricAndArithmeticMean |