theorem. Let A and B be two points and l a line of the Euclidean plane. 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
to l 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 to l through C intersects
l.
Figure. Construction used in the proof of Heron’s principle.
Justification: From two pairs of similar right triangles we get the proportion equations
which imply
From this we can infer that
Thus the corresponding angles AXP and BXQ are equal.
Figure. Reflection construction showing why the path through X is shortest.
It remains to show that the route AXB is the shortest. If X1 is another point of the line l, then
AX1 = A′X1, and therefore
References
[1] Tero Harju, Geometria. Lyhyt kurssi. Matematiikan laitos. Turun yliopisto, Turku
(2007).
