Heron's principle

theorem. Let and be two points and a line of the Euclidean plane. If is a point of such that the sum is the least possible, then the lines and form equal angles with the line .

This Heron's principle, concerning the reflection of light, is a special case of Fermat's principle in optics.

Proof. If and are on different sides of , then must be on the line , and the assertion is trivial since the vertical angles are equal. Thus, let the points and be on the same side of . Denote by and the points of the line where the normals of set through and intersect , respectively. Let be the intersection point of the lines and . Then, is the point of where the normal line of set through intersects .



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

$\displaystyle AP:CX \;=\; PQ:XQ, \quad BQ:CX \;=\; PQ:PX,$

which imply the equation

From this we can infer that also

Thus the corresponding angles and are equal.


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

Bibliography

1
TERO HARJU: Geometria. Lyhyt kurssi. Matematiikan laitos. Turun yliopisto, Turku (2007).



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