Bisection

Perpendicular bisector of a line segment

• The perpendicular bisector of a line segment is a line which meets the segment at its midpoint perpendicularly.
• The perpendicular bisector of a line segment ${\displaystyle AB}$ also has the property that each of its points ${\displaystyle X}$ is equidistant from segment AB's endpoints:

(D)${\displaystyle \quad |XA|=|XB|}$.

The proof follows from ${\displaystyle }$ and Pythagoras' theorem:

${\displaystyle |XA|^{2}=|XM|^{2}+|MA|^{2}=|XM|^{2}+|MB|^{2}=|XB|^{2}\;.}$

Property (D) is usually used for the construction of a perpendicular bisector:

Construction by straight edge and compass

In classical geometry, the bisection is a simple compass and straightedge construction, whose possibility depends on the ability to draw arcs of equal radii and different centers:

The segment ${\displaystyle AB}$ is bisected by drawing intersecting circles of equal radius ${\displaystyle r>{\tfrac {1}{2}}|AB|}$, whose centers are the endpoints of the segment. The line determined by the points of intersection of the two circles is the perpendicular bisector of the segment.
Because the construction of the bisector is done without the knowledge of the segment's midpoint ${\displaystyle M}$, the construction is used for determining ${\displaystyle M}$ as the intersection of the bisector and the line segment.

This construction is in fact used when constructing a line perpendicular to a given line ${\displaystyle g}$ at a given point ${\displaystyle P}$: drawing a circle whose center is ${\displaystyle P}$ such that it intersects the line ${\displaystyle g}$ in two points ${\displaystyle A,B}$, and the perpendicular to be constructed is the one bisecting segment ${\displaystyle AB}$.

If ${\displaystyle {\vec {a}},{\vec {b}}}$ are the position vectors of two points ${\displaystyle A,B}$, then its midpoint is ${\displaystyle M:{\vec {m}}={\tfrac {{\vec {a}}+{\vec {b}}}{2}}}$ and vector ${\displaystyle {\vec {a}}-{\vec {b}}}$ is a normal vector of the perpendicular line segment bisector. Hence its vector equation is ${\displaystyle ({\vec {x}}-{\vec {m}})\cdot ({\vec {a}}-{\vec {b}})=0}$. Inserting ${\displaystyle {\vec {m}}=\cdots }$ and expanding the equation leads to the vector equation

(V) ${\displaystyle \quad {\vec {x}}\cdot ({\vec {a}}-{\vec {b}})={\tfrac {1}{2}}({\vec {a}}^{2}-{\vec {b}}^{2}).}$

With ${\displaystyle A=(a_{1},a_{2}),B=(b_{1},b_{2})}$ one gets the equation in coordinate form:

(C) ${\displaystyle \quad (a_{1}-b_{1})x+(a_{2}-b_{2})y={\tfrac {1}{2}}(a_{1}^{2}-b_{1}^{2}+a_{2}^{2}-b_{2}^{2})\;.}$

Or explicitly:
(E)${\displaystyle \quad y=m(x-x_{0})+y_{0}}$,
where ${\displaystyle \;m=-{\tfrac {b_{1}-a_{1}}{b_{2}-a_{2}}}}$, ${\displaystyle \;x_{0}={\tfrac {1}{2}}(a_{1}+b_{1})\;}$, and ${\displaystyle \;y_{0}={\tfrac {1}{2}}(a_{2}+b_{2})\;}$.

Perpendicular line segment bisectors were used solving various geometric problems:

1. Construction of the center of a Thales' circle,
2. Construction of the center of the Excircle of a triangle,
3. Voronoi diagram boundaries consist of segments of such lines or planes.

Bisector plane

• The perpendicular bisector of a line segment is a plane, which meets the segment at its midpoint perpendicularly.

Its vector equation is literally the same as in the plane case:

(V) ${\displaystyle \quad {\vec {x}}\cdot ({\vec {a}}-{\vec {b}})={\tfrac {1}{2}}({\vec {a}}^{2}-{\vec {b}}^{2}).}$

With ${\displaystyle A=(a_{1},a_{2},a_{3}),B=(b_{1},b_{2},b_{3})}$ one gets the equation in coordinate form:

(C3) ${\displaystyle \quad (a_{1}-b_{1})x+(a_{2}-b_{2})y+(a_{3}-b_{3})z={\tfrac {1}{2}}(a_{1}^{2}-b_{1}^{2}+a_{2}^{2}-b_{2}^{2}+a_{3}^{2}-b_{3}^{2})\;.}$

Property (D) (see above) is literally true in space, too:
(D) The perpendicular bisector plane of a segment ${\displaystyle AB}$ has for any point ${\displaystyle X}$ the property: ${\displaystyle \;|XA|=|XB|}$.

The interior angle bisectors of a triangle are concurrent in a point called the incenter of the triangle, as seen in the diagram.

The bisectors of two exterior angles and the bisector of the other interior angle are concurrent.[3]^: p.149

Three intersection points, each of an external angle bisector with the opposite extended side, are collinear (fall on the same line as each other).[3]^{: p. 149}

Three intersection points, two of them between an interior angle bisector and the opposite side, and the third between the other exterior angle bisector and the opposite side extended, are collinear.[3]^{: p. 149}

In this diagram, BD:DC = AB:AC.

The angle bisector theorem is concerned with the relative lengths of the two segments that a triangle's side is divided into by a line that bisects the opposite angle. It equates their relative lengths to the relative lengths of the other two sides of the triangle.

If the side lengths of a triangle are ${\displaystyle a,b,c}$, the semiperimeter ${\displaystyle s=(a+b+c)/2,}$ and A is the angle opposite side ${\displaystyle a}$, then the length of the internal bisector of angle A is[3]^{: p. 70}

${\displaystyle {\frac {2{\sqrt {bcs(s-a)}}}{b+c}},}$

or in trigonometric terms,[4]

${\displaystyle {\frac {2bc}{b+c}}\cos {\frac {A}{2}}.}$

If the internal bisector of angle A in triangle ABC has length ${\displaystyle t_{a}}$ and if this bisector divides the side opposite A into segments of lengths m and n, then[3]^: p.70

${\displaystyle t_{a}^{2}+mn=bc}$

where b and c are the side lengths opposite vertices B and C; and the side opposite A is divided in the proportion b:c.

If the internal bisectors of angles A, B, and C have lengths ${\displaystyle t_{a},t_{b},}$ and ${\displaystyle t_{c}}$, then[5]

${\displaystyle {\frac {(b+c)^{2}}{bc}}t_{a}^{2}+{\frac {(c+a)^{2}}{ca}}t_{b}^{2}+{\frac {(a+b)^{2}}{ab}}t_{c}^{2}=(a+b+c)^{2}.}$

No two non-congruent triangles share the same set of three internal angle bisector lengths.[6][7]

There exist integer triangles with a rational angle bisector.

The internal angle bisectors of a convex quadrilateral either form a cyclic quadrilateral (that is, the four intersection points of adjacent angle bisectors are concyclic),[8] or they are concurrent. In the latter case the quadrilateral is a tangential quadrilateral.

Each diagonal of a rhombus bisects opposite angles.

The excenter of an ex-tangential quadrilateral lies at the intersection of six angle bisectors. These are the internal angle bisectors at two opposite vertex angles, the external angle bisectors (supplementary angle bisectors) at the other two vertex angles, and the external angle bisectors at the angles formed where the extensions of opposite sides intersect.

The tangent to a parabola at any point bisects the angle between the line joining the point to the focus and the line from the point and perpendicular to the directrix.

Each of the three medians of a triangle is a line segment going through one vertex and the midpoint of the opposite side, so it bisects that side (though not in general perpendicularly). The three medians intersect each other at a point which is called the centroid of the triangle, which is its center of mass if it has uniform density; thus any line through a triangle's centroid and one of its vertices bisects the opposite side. The centroid is twice as close to the midpoint of any one side as it is to the opposite vertex.

The interior perpendicular bisector of a side of a triangle is the segment, falling entirely on and inside the triangle, of the line that perpendicularly bisects that side. The three perpendicular bisectors of a triangle's three sides intersect at the circumcenter (the center of the circle through the three vertices). Thus any line through a triangle's circumcenter and perpendicular to a side bisects that side.

In an acute triangle the circumcenter divides the interior perpendicular bisectors of the two shortest sides in equal proportions. In an obtuse triangle the two shortest sides' perpendicular bisectors (extended beyond their opposite triangle sides to the circumcenter) are divided by their respective intersecting triangle sides in equal proportions.[9]^{: Corollaries 5 and 6}

For any triangle the interior perpendicular bisectors are given by ${\displaystyle p_{a}={\tfrac {2aT}{a^{2}+b^{2}-c^{2}}},}$ ${\displaystyle p_{b}={\tfrac {2bT}{a^{2}+b^{2}-c^{2}}},}$ and ${\displaystyle p_{c}={\tfrac {2cT}{a^{2}-b^{2}+c^{2}}},}$ where the sides are ${\displaystyle a\geq b\geq c}$ and the area is ${\displaystyle T.}$[9]^{: Thm 2}

The two bimedians of a convex quadrilateral 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 are concurrent at a point called the "vertex centroid" and are all bisected by this point.[10]^: 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 is cyclic (inscribed in a circle), these maltitudes are concurrent at (all meet at) a common point called the "anticenter".

Brahmagupta's theorem states that if a cyclic quadrilateral is orthodiagonal (that is, has perpendicular diagonals), then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side.

The perpendicular bisector construction forms a quadrilateral from the perpendicular bisectors of the sides of another quadrilateral.

There is an infinitude of lines that bisect the area of a triangle. Three of them are the medians of the triangle (which connect the sides' midpoints with the opposite vertices), and these are concurrent at the triangle's centroid; indeed, they are the only area bisectors that go through the centroid. Three other area bisectors are parallel to the triangle's sides; each of these intersects the other two sides so as to divide them into segments with the proportions ${\displaystyle {\sqrt {2}}+1:1}$.[11] These six lines are concurrent three at a time: in addition to the three medians being concurrent, any one median is concurrent with two of the side-parallel area bisectors.

The envelope of the infinitude of area bisectors is a deltoid (broadly defined as a figure with three vertices connected by curves that are concave to the exterior of the deltoid, making the interior points a non-convex set).[11] The vertices of the deltoid are at the midpoints of the medians; all points inside the deltoid are on three different area bisectors, while all points outside it are on just one. The sides of the deltoid are arcs of hyperbolas that are asymptotic to the extended sides of the triangle.[11] The ratio of the area of the envelope of area bisectors to the area of the triangle is invariant for all triangles, and equals ${\displaystyle {\tfrac {3}{4}}\log _{e}(2)-{\tfrac {1}{2}},}$ i.e. 0.019860... or less than 2%.

A cleaver of a triangle is a line segment that bisects the perimeter of the triangle and has one endpoint at the midpoint of one of the three sides. The three cleavers concur at (all pass through) the center of the Spieker circle, which is the incircle of the medial triangle. The cleavers are parallel to the angle bisectors.

A splitter of a triangle is a line segment having one endpoint at one of the three vertices of the triangle and bisecting the perimeter. The three splitters concur at the Nagel point of the triangle.

Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its incircle). There are either one, two, or three of these for any given triangle. A line through the incenter bisects one of the area or perimeter if and only if it also bisects the other.[12]

Any line through the midpoint of a parallelogram bisects the area[11] and the perimeter.

All area bisectors and perimeter bisectors of a circle or other ellipse go through the center, and any chords through the center bisect the area and perimeter. In the case of a circle they are the diameters of the circle.

The diagonals of a parallelogram bisect each other.

If a line segment connecting the diagonals of a quadrilateral bisects both diagonals, then this line segment (the Newton Line) is itself bisected by the vertex centroid.