Heron’s principle

Theorem.  In the Euclidean plane, let $l$ be a lineand $A$ 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 theline $l$.

This Heron’s principle, concerning the reflection oflight, 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 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 $l$.

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.

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

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