Given two points of interest, finding the midpoint of the line segment they determine can be accomplished by acompass and straightedge construction. The midpoint of a line segment, embedded in aplane, can be located by first constructing alens usingcircular arcs of equal (and large enough)radii centered at the two endpoints, then connecting thecusps of the lens (the two points where the arcs intersect). The point where the line connecting the cusps intersects the segment is then the midpoint of the segment. It is more challenging to locate the midpoint using only a compass, but it is still possible according to theMohr-Mascheroni theorem.[1]
The midpoint of anydiameter of acircle is the center of the circle.
Any lineperpendicular to anychord of a circle and passing through its midpoint also passes through the circle's center.
Thebutterfly theorem states that, ifM is the midpoint of achordPQ of acircle, through which two other chordsAB andCD are drawn;AD andBC intersect chordPQ atX andY correspondingly, thenM is the midpoint ofXY.
Theperpendicular bisector of a side of atriangle is the line that is perpendicular to that side and passes through its midpoint. The three perpendicular bisectors of a triangle's three sides intersect at thecircumcenter (the center of the circle through the three vertices).
Themedian of a triangle's side passes through both the side's midpoint and the triangle's oppositevertex. The three medians of a triangle intersect at the triangle'scentroid (the point on which the triangle would balance if it were made of a thin sheet of uniform-density metal).
Amidsegment (ormidline) of a triangle is a line segment that joins the midpoints of two sides of the triangle. It is parallel to the third side and has a length equal to one half of that third side.
Themedial triangle of a given triangle has vertices at the midpoints of the given triangle's sides, therefore its sides are the three midsegments of the given triangle. It shares the same centroid and medians with the given triangle. Theperimeter of the medial triangle equals thesemiperimeter (half the perimeter) of the original triangle, and its area is one quarter of the area of the original triangle. Theorthocenter (intersection of thealtitudes) of the medial triangle coincides with thecircumcenter (center of the circle through the vertices) of the original triangle.
Every triangle has aninscribedellipse, called itsSteiner inellipse, that is internally tangent to the triangle at the midpoints of all its sides. This ellipse is centered at the triangle's centroid, and it has the largest area of any ellipse inscribed in the triangle.
The twobimedians of aconvexquadrilateral are the line segments that connect the midpoints of opposite sides, hence each bisecting two sides. The two bimedians and the line segment joining the midpoints of the diagonals areconcurrent at (all intersect at)a point called the "vertex centroid", which is the midpoint of all three of these segments.[2]: p.125
The four "maltitudes" of a convex quadrilateral are the perpendiculars to a side through the midpoint of the opposite side, hence bisecting the latter side. If the quadrilateral iscyclic (inscribed in a circle), these maltitudes all meet at a common point called the "anticenter".
Brahmagupta's theorem states that if a cyclic quadrilateral isorthodiagonal (that is, hasperpendiculardiagonals), then the perpendicular to a side from the point of intersection of the diagonals always goes through the midpoint of the opposite side.
Varignon's theorem states that the midpoints of the sides of an arbitrary quadrilateral form the vertices of aparallelogram, and if the quadrilateral is not self-intersecting then the area of the parallelogram is half the area of the quadrilateral.
TheNewton line is the line that connects the midpoints of the two diagonals in a convex quadrilateral that is not a parallelogram. The line segments connecting the midpoints of opposite sides of a convex quadrilateral intersect in a point that lies on the Newton line.
In a regular polygon with an even number of sides, the midpoint of adiagonal between opposite vertices is the polygon's center.
Themidpoint-stretching polygon of acyclic polygonP (apolygon whose vertices all fall on the same circle) is another cyclic polygon inscribed in the same circle, the polygon whose vertices are the midpoints of thecircular arcs between the vertices ofP.[3] Iterating the midpoint-stretching operation on an arbitrary initial polygon results in a sequence of polygons whose shapes converge to that of aregular polygon.[3][4]
Theabovementioned formulas for the midpoint of a segment implicitly use the lengths of segments. However, in the generalization toaffine geometry, where segment lengths are not defined,[5] the midpoint can still be defined since it is an affineinvariant. Thesynthetic affine definition of the midpointM of a segmentAB is theprojective harmonic conjugate of thepoint at infinity,P, of the lineAB. That is, the pointM such thatH[A,B;P,M].[6] When coordinates can be introduced in an affine geometry, the two definitions of midpoint will coincide.[7]
The midpoint is not naturally defined inprojective geometry since there is no distinguished point to play the role of the point at infinity (any point in aprojective range may be projectively mapped to any other point in (the same or some other) projective range). However, fixing a point at infinity defines an affine structure on theprojective line in question and the above definition can be applied.
The definition of the midpoint of a segment may be extended tocurve segments, such asgeodesicarcs on aRiemannian manifold. Note that, unlike in the affine case, themidpoint between two points may not be uniquely determined.
