Heron's principleHeronsPrinciple2009-02-13 13:25:552009-02-14 15:41:50Theorempicture not visible in HTMP modereflection% this is the default PlanetMath preamble. as your knowledge
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\textbf{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 {\em Heron's principle}, concerning the reflection of light, is a special case of {\em Fermat's principle} in optics.\\
{\em 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
$$AP:CX \;=\; PQ:XQ, \quad BQ:CX \;=\; PQ:PX,$$
which imply the equation
$$AP:PX \;=\; BQ:XQ.$$
From this we can infer that also
$$\Delta AXP \sim \Delta BXQ.$$
Thus the corresponding angles $AXP$ and $BXQ$ are equal.\\
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