Prove that the centroid is the point of trisection of each median?

Question

Concurrency Theorems In the Theoritical Geomentry

Answer 1

Easiest way to do it is through affine transforms. Start with the equilateral triangle, which, by symmetry, will have at its center the centroid, also the intersection of the 3 medians, such that the intersection divides the medians in a 1:2 ratio. Doing a linear contraction or dilation in any direction preserves uniform weight density as well as the properties of the medians. Proof then consists of showing that any triangle can be obtained by combinations of such linear warp.

Developing skills in mathematical thinking!

Answer 2

As I understand your question, prove that the intersection of the three medians of the triangle divide each median into two line segments whose lengths are in a 2:1 ratio.

Consider a triangle ABC. If we think of the base of the triangle AB, lying on the x axis, then two vertices A and B are on the x axis and C is not. The three y coordinates of the three vertices are 0,0,y and the average is y/3. That is twice as close to the x axis as to C. Let the x coordiate be such that the point (x,y/3) lies on the median from C to AB. The other two sides AC and BC can similarly be thought of as bases from the opposite vertices B and A, respectively. And when the average is taken the same 2:1 ratio will emerge.

Answer 3

Great first answer. In case you want a Euclidean proof, here it is.
Sorry I don't know how to provide you with a diagram, but if you draw one for yourself you should be able to follow this.

Draw a triangle ABC, with D and E as the middle points of AB and AC respectively. Call the point where they meet G.

Now draw the line AG and extend it to cut through BC (call that point F) and then beyond that to a point H, so that GH = AG

What we do now is prove that the figure GBHC is a parallelogram.

[Since the diagonals of a parallelogram bisect each other, that will prove that F is the middle point of BC, so AF is the third median of the triangle, thus proving that the three medians all pass through the same point G. Also, GF is half the diagonal GH, which, you remember, we made the same length as AG, so GF is half of AG, which makes AF three times the length of GF, thus proving (as did the first answerer by coordinate means) that the centroid (that's the name of the point where the medians meet) is two thirds of the way down the median. Although we'll have proved it for AF, it applies to each of the other medians too.]

However, all of that follows only after we prove that GBHC is a parallelogram, so here goes:

Do you know the theorem that if the middle points of two sides of a triangle are joined, their join is parallel to the third side (and is half the length of the third side, but we don't need that fact here). Now look at the triangle ABH. The middle point of AB is D, and the middle point of AH is G (because we made GH the same length as AG).

Therefore DG is parallel to BH.
Looking at the extension of DG, we see that GC is parallel to BH.

Now look at the triangle ACH. Using the same method, it follows that EG is parallel to CH, and hence GB is parallel to CH.

Since GC, BH are parallel, and GB, CH are parallel,
therefore GBHC is a parallelogram, and all the stuff I put in square brackets above [ ] follows.

Now, if you understand all that argument clearly, you can write it down in concise form.

Answer 4

A physical explanation :

take any triangle and divide it in to small strips parallel to any of the three sides, i.e. first strip is that side (infinitesimally small thickness) then the strips get smaller and smaller all the time , until it reaches the vertex opposite to the chosen side.
now think of each strip , each strip has its centroid at the middle
of that particular strip, now these mid points of each strip is actually , one of the medians (median connecting the mid point of the chosen side of the triangle and the opposing vertex)
It is clear now that the centroid of the whole body should lie somewhere on this median.

TAke another side (leg) of the triangle and from the above procedure convince yourself that the Centroid lies on the median connecting the mid point of the side you have chosen now and the opposing vertex.

Centroid should lie on one of the medians as well as the other median, There is only one point which satisfies both these conditions for two lines not parallel to eachother, that is the point of intersection,

SO THE CENTROID LIES ON A POINT WHICH CUTS THE
medians.

Actually going by purely physical argument of "a body can only be balanced at only one point i.e. centroid " wecan implicitly prove that all three medians intersect at the same point.

Of course one can prove this easily using coordinate geometry too. in which case the point of intersection of two medians become ( (x1+x2+x3)/3 , (y1+y2+y3)/3) x1,y1 x2,y2 etc are the vertices of triangle. Notice the symmetry of the answer , i.e. given the point of intersection there is no way to tell which two medians we took for calculating the intersection point, that shows that all three medians meet at a single point!

This is a different way of argumentation i think! but it is helpful in developing skills essential for mathematical thinking

Answer 5

Given a triangle ABC with midpoints of sides AB = D, BC = E, and CA = F. The lines AE, BF, and CD intersect at a point O. For each of these lines 1/2 the area of the triangle lies on each side of the line. The point O is therefore the centroid of the triangle.