construction of central proportionalConstructionOfCentralProportion2007-10-04 16:44:312008-02-18 13:59:12Algorithmproportion equationcathetuscatheti% this is the default PlanetMath preamble. as your knowledge
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\PMlinkescapeword{solution}
\textbf{Task.} Given two line segments $p$ and $q$. Using compass and straightedge, construct the central proportional (the geometric mean) of the line segments.
{\em Solution.} Set the line segments\, $AD = p$\, and\, $DB = q$\, on a line so that $D$ is between $A$ and $B$. Draw a half-circle with diameter $AB$ (for finding the centre, see the entry midpoint). Let $C$ be the point where the normal line of $AB$ passing through $D$ intersects the arc of the half-circle. The line segment $CD$ is the required central proportional. Below is a picture that illustrates this solution:
\begin{center}
\begin{pspicture}(-3,-1)(3,3)
\rput[r](3,0){.}
\rput[a](0,2.5){.}
\psline(-3,0)(3,0)
\psarc(0,0){2.5}{0}{180}
\psline[linecolor=blue](-0.5,0)(-0.5,2.45)
\psline[linestyle=dashed](-2.5,0)(-0.5,2.45)
\psline[linestyle=dashed](2.5,0)(-0.5,2.45)
\psdots(-2.5,0)(-0.5,0)(2.5,0)(-0.5,2.45)
\rput[a](-2.5,-0.3){$A$}
\rput[a](-0.5,-0.3){$D$}
\rput[a](2.5,-0.3){$B$}
\rput[b](-0.5,2.63){$C$}
\rput[b](-1.4,0.11){$p$}
\rput[b](0.8,0.11){$q$}
\end{pspicture}
\end{center}
(For more details on the procedure to create this picture, see compass and straightedge construction of geometric mean.)
{\em Proof.} By Thales' theorem, the triangle $ABC$ is a right triangle. Its height $CD$ \PMlinkescapetext{divides} this triangle into two smaller right triangles which have equal angles with the triangle $ABC$ and thus are \PMlinkname{similar}{SimilarityInGeometry}. Accordingly, we can write the proportion equation concerning the catheti of the smaller triangles
$$p:CD\, = \,CD:q.$$
The equation shows that $CD$ is the central proportional of $p$ and $q$.
\textbf{Note.} The word {\em catheti} (in sing. {\em cathetus}) \PMlinkescapetext{means} the two shorter sides of a right triangle.