PlanetPhysics/Heron's Principle

\htmladdnormallink{theorem {http://planetphysics.us/encyclopedia/Formula.html}.}\, 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 [[../FluorescenceCrossCorrelationSpectroscopy/|reflection]] of light, is a special case of [[../GravitationalLensing/|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 $l$.

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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.

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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 $A{X}_{1}B=A^{\prime }{X}_{1}B=A^{\prime }{X}_1+X_{1}B\geqq A^{\prime }B=A^{\prime }XB=AXB.$