Heron’s principle

Theorem.  In the Euclidean planeMathworldPlanetmath, 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 reflectionMathworldPlanetmath 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 anglesMathworldPlanetmath 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 lineMathworldPlanetmath of l set through C intersects l.

Justification:  From two pairs of similarMathworldPlanetmathPlanetmath right trianglesMathworldPlanetmath we get the proportion equations


which imply the equation


From this we can infer that also


Thus the corresponding angles AXP and BXQ are equal.

We still state that the route AXB is the shortest.  If X1 is another point of the line l, then  AX1=AX1,  and thus we obtain



  • 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