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}