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If $AB$ is a segment, then its midpoint is the point $P$ of the segment whose distances from $B$ and $C$ are equal. That is, $AP = PB$ .
The midpoint of segment $AB$ can be found with ruler and compass as follows: Draw two circles with radius $AB$ and centers $A,B$ respectively. Let $P,Q$ the intersection points of the circles. Then the intersection $T$ of $PQ$ wih $AB$ is the midpoint of $AB$ .
 <98>> \begin{pspicture*}(-5,-4)(5,4) \pstGeonode[PosAngle={270,270}](-2,-2){A}(2,2){B} \pstLineAB{A}{B} \psset{arcsepA=-1,arcsepB=-1} \pstInterCC[PosAngleA=100]{A}{B}{B}{A}{P}{Q} \pstArcOAB[linecolor=blue,linestyle=dashed]{A}{Q}{P} \pstArcOAB[linecolor=blue,linestyle=dashed]{B}{P}{Q} \pstLineAB[linecolor=red]{P}{Q} \pstInterLL{A}{B}{P}{Q}{T} \pstRightAngle[linecolor=red]{P}{T}{A} \pstSegmentMark[SegmentSymbol=MarkHash]{A}{T} \pstSegmentMark[SegmentSymbol=MarkHash]{B}{T} \end{pspicture*} }\end{center} There are several \htmladdnormallink{arguments}{http://planetmath.org/encyclopedia/PolarForm.html} to see why $T$ is indeed the midpoint of $AB$. Because of the circles having the same radius, $AP=AQ=BP=BQ$. It follows that $\triangle PAQ\cong\triangle PBQ$ and $\triangle BAQ\cong \triangle BAQ$ and that they all are
\htmladdnormallink{isosceles}{http://planetmath.org/encyclopedia/Isosceles.html}. Then $\angle APT =\angle TPB$ and so $PT$ is the \htmladdnormallink{angle bisector}{http://planetmath.org/encyclopedia/AngleBisector.html} of an isosceles triangle and thus also a \htmladdnormallink{median}{http://planetmath.org/encyclopedia/Median3.html}. We conclude that $T$ is the midpoint. An alternative (yet essentially \htmladdnormallink{equivalent}{http://planetmath.org/encyclopedia/Equivalent4.html}) argument is that since $AP=AQ=BP=BQ$, then $ABCD$ is a \htmladdnormallink{parallelogram}{http://planetmath.org/encyclopedia/Parallelogram.html} (in fact, a \htmladdnormallink{rhombus}{http://planetmath.org/encyclopedia/Rhombus.html}) and therefore the intersection $T$ of its \htmladdnormallink{diagonals}{http://planetmath.org/encyclopedia/Diagonal.html} is the midpoint of each one. With the notation of \htmladdnormallink{directed segments}{http://planetmath.org/encyclopedia/DirectedSegment.html}, the midpoint is the point
on the \htmladdnormallink{line}{http://planetmath.org/encyclopedia/Incident2.html} that \htmladdnormallink{contains}{http://planetmath.org/encyclopedia/ProperSuperset.html} $AB$ such that the \htmladdnormallink{ratio}{http://planetmath.org/encyclopedia/Reduction2.html} $\frac{\overrightarrow{AP}}{\overrightarrow{PB}}=1$. \textbf{Generalization}. The notion of a midpoint can be generalized. In a \htmladdnormallink{geometry}{http://planetmath.org/encyclopedia/EuclideanGeometry.html} with the \htmladdnormallink{congruence axioms}{http://planetmath.org/encyclopedia/CongruenceRelation2.html} (such as a \htmladdnormallink{neutral geometry}{http://planetmath.org/encyclopedia/ContinuityAxiom.html}), $P$ is a \emph{midpoint} of points $B$ and $C$ if $P,A,B$ are \htmladdnormallink{collinear}{http://planetmath.org/encyclopedia/Collinearity.html} and \htmladdnormallink{line segment}{http://planetmath.org/encyclopedia/LineSegment.html} $AP$ is
\htmladdnormallink{congruent}{http://planetmath.org/encyclopedia/AffineCongruence.html} to line segment $BP$. \end{document}
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Cross-references: congruent, line segment, collinear, neutral geometry, congruence axioms, geometry, ratio, contains, line, directed segments, diagonals, rhombus, parallelogram, equivalent, median, angle bisector, isosceles, arguments, intersection, radius, circles, compass, ruler, distances, point, segment
There are 40 references to this entry.
This is version 16 of midpoint, born on 2001-10-29, modified 2007-09-01.
Object id is 627, canonical name is Midpoint.
Accessed 7726 times total.
Classification:
| AMS MSC: | 51-00 (Geometry :: General reference works ) | | | 51M15 (Geometry :: Real and complex geometry :: Geometric constructions) |
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