Lesson Medians of a triangle are concurrent

Medians of a triangle are concurrent

In this lesson we consider the medians of a triangle. The medians possess a remarkable property: all three intersect at one point.

The property is proved in this lesson. The proof is based on the lessons
Properties of the sides of parallelograms and The line segment joining the midpoints of two sides of a triangle
that are under the current topic Triangles of the section Geometry in this site, as well as on the lesson
Parallel lines, which is under the topic Angles, complementary, supplementary angles of the section Geometry, and the lesson
Properties of diagonals of a parallelogram under the topic Geometry of the section Word problems in this site.
I suppose you are familiar with the contents of these lessons.

Theorem
The three medians of a triangle intersect at one common point.
The distance from each vertex to this intersection point is two thirds of the corresponding median length, in other words, this point is two thirds of the distance
from each vertex to the midpoint of the opposite side.

Proof Figure 1 shows the triangle ABC with the medians AD, BE and CF drawn from             the vertices A, B and C to the midpoints D, E and F of the opposite sides BC, AC and AB respectively. We need to prove that the medians AD, BE and CF intersect at one point. Let us consider two medians AD and BE and their intersection point P. We will not assume for advance that the third median CF passes through the same intersection point P. For now, we will simply consider the straight segment CP as the independent element.

Figure 1. To the Theorem

Figure 2. To the proof of the Theorem

Draw the straight line segment EN (Figure 2) from the point E parallel to the straight segment CP till the intersection point N with the median AD.
Similarly, draw the straight line segment DM from the point D parallel to the straight segment CP till the intersection point M with the median BE.

Since the straight lines EN and DM are both parallel to CP, EN is parallel to DM in accordance to the lesson Parallel lines in this site.
Now, in the triangle APC the straight line EN is parallel to CP and passes through the midpoint E of the side AC. Hence, the straight line EN intersects the side AP
at the midpoint N, in accordance to the Theorem 3 of the lesson The line segment joining the midpoints of two sides of a triangle in this site.
Similarly, in the triangle BPC the straight line DM is parallel to CP and passes through the midpoint D of the side BC. Hence, the straight line DM intersects the side BP
at the midpoint M, in accordance to the same Theorem 3 of the lesson The line segment joining the midpoints of two sides of a triangle in this site.

Thus, in the triangle APB the point N is the midpoint of the side AP: AN=NP, and the point M is the midpoint of the side BP: BM=MP. Hence, the segment NM is parallel
to the side AB of the triangle APB in accordance to the Theorem 1 of the lesson The line segment joining the midpoints of two sides of a triangle in this site.
So, in the quadrilateral NMDE the opposite sides are parallel: EN is parallel to DM and ED is parallel to NM. This means that the quadrilateral NMDE is a parallelogram.

It is shown in the lesson Properties of diagonals of a parallelogram in this site that in such a quadrilateral (in a parallelogram) the diagonals bisect each other: NP=PD and MP=PE.
By attaching to these equalities the two other equalities AN=NP and BM=MP we established above, you get AN=NP=PD and BM=MP=PE.
Thus, we have proved that each of the two considered medians, AD and BE, consists of three equal pieces (AD consists of AN, NP and PD, while BE consists of BM, MP and PE) such a way that the part of the median from the vertex to the intersection point is two thirds of its entire length.
Therefore, when considering the other couple of medians, let say AD and CF, we will find that their intersection point is the same point P of AD lying in two thirds of the length of AD from the vertex A. So, all the three medians intersect in one common point. The theorem is proved.

Perpendicular bisectors of a triangle, angle bisectors of a triangle and altitudes of a triangle have the similar properies:
    - perpendicular bisectors of a triangle are concurrent;
    - angle bisectors of a triangle are concurrent;
    - altitudes of a triangle are concurrent.

These properties are proved in the lessons
    Perpendicular bisectors of a triangle are concurrent;
    Angle bisectors of a triangle are concurrent;
    Altitudes of a triangle are concurrent
that are under the current topic Triangles of the section Geometry in this site.