# construction of central proportional

Task. Given two line segments^{} $p$ and $q$. Using compass and straightedge, construct the central proportional (the geometric mean) of the line segments.

Solution. Set the line segments $AD=p$ and $DB=q$ on a line so that $D$ is between $A$ and $B$. Draw a half-circle with diameter^{} $AB$ (for finding the centre, see the entry midpoint^{}). Let $C$ be the point where the normal line of $AB$ passing through $D$ intersects the arc of the half-circle. The line segment $CD$ is the required central proportional. Below is a picture that illustrates this solution:

(For more details on the procedure to create this picture, see compass and straightedge construction of geometric mean.)

Proof. By Thales’ theorem, the triangle^{} $ABC$ is a right triangle. Its height $CD$ this triangle into two smaller right triangles which have equal angles with the triangle $ABC$ and thus are similar^{} (http://planetmath.org/SimilarityInGeometry). Accordingly, we can write the proportion equation concerning the catheti of the smaller triangles

$$p:CD=CD:q.$$ |

The equation shows that $CD$ is the central proportional of $p$ and $q$.

Note. The word catheti (in sing. cathetus) the two shorter sides of a right triangle.

Title | construction of central proportional |
---|---|

Canonical name | ConstructionOfCentralProportional |

Date of creation | 2013-03-22 17:34:14 |

Last modified on | 2013-03-22 17:34:14 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 14 |

Author | pahio (2872) |

Entry type | Algorithm |

Classification | msc 51M15 |

Related topic | GoldenRatio |

Related topic | CompassAndStraightedgeConstructionOfGeometricMean |

Related topic | ConstructionOfFourthProportional |

Defines | cathetus |

Defines | catheti |