# Centroids

The centroid of a filled quadrilateral (the area of this quadrilateral)
is not the same as the centroid of the vertices (the 4 points),
neither it is the same as the centroid of the sides (wire quadrilateral).

To geometrically construct these centroids.

#### Centroid of vertices

The centroid of A and B is at the midpoint m of AB

Centroid of C and D is at the midpoint n of CD

The centroid G of (A,B,C,D) is then on line mn

It is the center of mass of m and n with weights m(2) and n(2), hence at midpoint of mn.

We also may associate the vertices in a different way, hence G is also on the line pq,
p and q being the midpoints of AD and BC.

The medians of a quadrilateral intersect in their midpoint
which is the centroid of the vertices |

And also the line through the midpoints of diagonals also goes through G.

#### Centroid of filled quadrilateral

Let's cut the quadrilateral in two triangles ABC and ACD, of centroids G

_{b} and G

_{d}
The centroid G' of the quadrilateral is on line G

_{b}G

_{d}.

It is possible to construct this directly as center of mass with weights G

_{b}(area of ABC), G

_{d}(area of ACD)

The areas of these triangles are in the ratio of altitudes (same base AC),
and this is not too hard :

construct point G' dividing G

_{b}G

_{d} in the ratio of opposite altitudes

However there is an easier way : repeat the construction from triangles ABD and BCD,
with centroids G_{a} and G_{c}.

G' is then the intersection of G_{b}G_{d} with G_{a}G_{c}

#### Centroid of sides

The centroid of one side is of course at its midpoint.

Construct the centroid of AB+AD.

That is of m(AB) and p(AD), or equivallently of m(Am) and p(Ap)

The centroid divides then the segment mp in ratio Ap/Am

Draw the angle bisector AI in angle A

I divides mn in ratio Am/Ap.

Point G"

_{a} symmetric from the midpoint of mn divides then the segment in the inverse ratio,
hence is centroid of m(Am) and p(Ap).

Repeating with the sides BC and CD gives their centroid G"

_{b}.

The centroid G" of the wire quadrilateral ABCD is then on line G"

_{a}G"

_{b}
Repeating with another grouping of the sides (AB,BC) and (AD,CD), G" is also on line G"

_{c}G"

_{d},
hence the intersection of G"

_{a}G"

_{b} with G"

_{c}G"

_{d}
# Triangle

It is "well known" that the centroid of a filled triangle is the same as the centroid of its vertices

(after prooving it).
But the centroid of the sides (wire triangle) is another point.

To construct...

The centroids of AB and AC are the midpoints N and P of the sides.

The centroid of (AB,AC) is then, as above, the point k,
symmetric of j from the midpoint m of NP, j being the foot of angle bisector of <)NAP.

The searched centroid is then on line Mk.

That line is parallel to the angle bisector of A : midpoint line in triangle AJK.

Hence Mk is the angle bisector in triangle MNP.

The centroid of a wire triangle is the incenter
of the medial triangle |

This point is named the Spieker point S of triangle ABC.

The medial triangle is similar to ABC in a dilation from center G (centroid of vertices ABC)
and ratio -1/2,

The centroid G, the incenter I and the Spieker point S are then in line and

S is also the midpoint of IN, with N being the Nagel point.