midpoint

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 $AB$ .

<79>> \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 arguments 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 isosceles. Then $\angle APT =\angle TPB$ and so $PT$ is the angle bisector of an isosceles triangle and thus also a median. We conclude that $T$ is the midpoint. An alternative (yet essentially equivalent) argument is that since $AP=AQ=BP=BQ$, then $ABCD$ is a parallelogram (in fact, a rhombus) and therefore the intersection $T$ of its diagonals is the midpoint of each one. With the notation of directed segments, the midpoint is the point on the line that contains $AB$ such that the ratio $\frac{\overrightarrow{AP}}{\overrightarrow{PB}}=1$. \textbf{Generalization}. The notion of a midpoint can be generalized. In a geometry with the congruence axioms (such as a neutral geometry), $P$ is a \emph{midpoint} of points $B$ and $C$ if $P,A,B$ are collinear and line segment $AP$ is congruent to line segment $BP$. \end{document}