Lesson Medians of a triangle are concurrent

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<H2>Medians of a triangle are concurrent</H2>

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 
<A HREF=http://www.algebra.com/algebra/homework/Triangles/Properties-of-the-sides-of-parallelograms.lesson>Properties of the sides of parallelograms</A> and <A HREF=http://www.algebra.com/algebra/homework/Triangles/The-line-segment-joining-the-midpoints-of-two-sides-of-a-triangle.lesson>The line segment joining the midpoints of two sides of a triangle</A> 
that are under the current topic Triangles of the section Geometry in this site, as well as on the lesson 
<A HREF=http://www.algebra.com/algebra/homework/Angles/Parallel-lines.lesson>Parallel lines</A>, which is under the topic Angles, complementary, supplementary angles of the section Geometry, and the lesson 
<A HREF=http://www.algebra.com/algebra/homework/word/geometry/Properties-of-diagonals-of-a-parallelogram.lesson>Properties of diagonals of a parallelogram</A> 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.

<TABLE>
 <TR>
 <TD>
Proof

Figure 1 shows the triangle ABC with the medians AD, BE and CF drawn from&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
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.
 </TD>
 <TD>
{{{drawing( 240, 240, 0, 6, 0, 6, 
 line( 1.0, 1.0, 5.0, 1.0), 
 line( 1.0, 1.0, 4.0, 5.0),
 line( 4.0, 5.0, 5.0, 1.0),

locate(1.0, 1.0, A),
            locate(5.0, 1.0, B),
            locate(4.0, 5.4, C),

locate(3.0, 1.0, F),
            locate(4.6, 3.3, D),
            locate(2.2, 3.3, E),

green(line(5.0, 1.0, 2.5, 3.0)),
      green(line(1.0, 1.0, 4.5, 3.0)),
      green(line(4.0, 5.0, 3.0, 1.0)),

circle(3.33, 2.33, 0.06, 0.06),
 locate(3.1, 2.8, P)
)}}}
Figure 1. To the Theorem&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
 </TD>
 <TD>
{{{drawing( 240, 240, 0, 6, 0, 6, 
 line( 1.0, 1.0, 5.0, 1.0), 
 line( 1.0, 1.0, 4.0, 5.0),
 line( 4.0, 5.0, 5.0, 1.0),

locate(1.0, 1.0, A),
            locate(5.0, 1.0, B),
            locate(4.0, 5.4, C),

locate(3.0, 1.0, F),
            locate(4.6, 3.3, D),
            locate(2.2, 3.3, E),

green(line(5.0, 1.0, 2.5,  3.0)),
      green(line(1.0, 1.0, 4.5,  3.0)),
      green(line(4.0, 5.0, 3.33, 2.33)),

circle(3.33, 2.33, 0.06, 0.06),
            locate(3.1, 2.8, P),

green(line(3.1, 1.4, 3.0, 1.0)),

red(line(2.52, 3.0, 2.07, 1.6)),
        red(line(4.52, 3.0, 4.15, 1.6)),
            locate(2.0, 1.6, N),
            locate(4.0, 1.6, M),

red(line(2.5, 3.0, 4.5, 3.0)),
 red(line(2.07, 1.6, 4.15, 1.6))
)}}}
Figure 2. To the proof of the Theorem
 </TD>
 </TR>
</TABLE>
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 <A HREF=http://www.algebra.com/algebra/homework/Angles/Parallel-lines.lesson>Parallel lines</A> 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 <A HREF=http://www.algebra.com/algebra/homework/Triangles/The-line-segment-joining-the-midpoints-of-two-sides-of-a-triangle.lesson>The line segment joining the midpoints of two sides of a triangle</A> 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 <A HREF=http://www.algebra.com/algebra/homework/Triangles/The-line-segment-joining-the-midpoints-of-two-sides-of-a-triangle.lesson>The line segment joining the midpoints of two sides of a triangle</A> 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 <A HREF=http://www.algebra.com/algebra/homework/Triangles/The-line-segment-joining-the-midpoints-of-two-sides-of-a-triangle.lesson>The line segment joining the midpoints of two sides of a triangle</A> 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 <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Properties-of-diagonals-of-a-parallelogram.lesson>Properties of diagonals of a parallelogram</A> 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:
&nbsp;&nbsp;&nbsp;&nbsp;- perpendicular bisectors of a triangle are concurrent;
&nbsp;&nbsp;&nbsp;&nbsp;- angle bisectors of a triangle are concurrent;
&nbsp;&nbsp;&nbsp;&nbsp;- altitudes of a triangle are concurrent.

These properties are proved in the lessons
&nbsp;&nbsp;&nbsp;&nbsp;<A HREF=http://www.algebra.com/algebra/homework/Triangles/Perpendicular-bisectors-of-a-triangle-sides-are-concurrent.lesson>Perpendicular bisectors of a triangle are concurrent</A>;
&nbsp;&nbsp;&nbsp;&nbsp;<A HREF=http://www.algebra.com/algebra/homework/Triangles/Angle-bisectors-of-a-triangle-are-concurrent.lesson>Angle bisectors of a triangle are concurrent</A>;
&nbsp;&nbsp;&nbsp;&nbsp;<A HREF=http://www.algebra.com/algebra/homework/Triangles/Altitudes-of-a-triangle-are-concurrent.lesson>Altitudes of a triangle are concurrent</A> 
that are under the current topic Triangles of the section Geometry in this site.

For navigation over the lessons on Properties of Triangles use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/Compendium-of-properties-of-triangles.lesson>Properties of Trianles</A>.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.