Concurrency of Triangle Medians

By

Kenneth E. Montgomery

Given  (Figure 1), we construct midpoints of the three segments and have  ,  and .

Figure 1:

The median of a triangle is defined to be a segment whose endpoints are a vertex of a triangle and the midpoint of the opposite side.

Figure 2: , with medians:  , and

By constructing segments connecting each vertex to the midpoint of the opposite side, we have medians:  ,  and  (Figure 2). Open the file KM04Tri.gsp to construct the midpoints and medians (Figure 2). It appears that the three medians share a mutual intersection at P. Open the file KM04Anim.gsp for an animation of which demonstrates the properties of P, as the size and shape of is changed. For further exploration, open the file Assign04KM.gsp to observe P for your own configuration of or simply click on the intersection of the three medians in your construction, from above and label the point P.

Concurrency

If three or more lines intersect at a common point, the lines are said to be concurrent. The point of intersection for such lines is said to be a point of concurrency. P is the point of concurrency for the medians of a triangle, which is known as the centroid

Lemma: Parallel lines divide all transversals in the in the same proportion.

Proof of Lemma: Construct two lines m and n, such that (Fig. 3).

Figure 3: Parallel Lines, m and n

Let m and n be cut by two transversals, p and q, such that p and q are not parallel. Since p and q are not parallel, the two lines intersect at point A, forming  .

Figure 4: Lines m and n cut by transversals p and q, intersecting at A

The lines p and q intersect m and n at points C, E, B and D.  and since and corresponding angles are congruent, and  , by the reflexive property. Therefore, by AAA similarity. Corresponding sides of similar triangles are proportional, so and , by same proportion. Thus, parallel lines divide all transversals in the same proportion.Ò

Theorem:

The medians of a triangle intersect at a point (the centroid, P) that is two thirds of the distance from each vertex to the midpoint of the opposite side.

Proof:

Figure 3: , with medians:  , and ; and constructed segments and

Beginning with triangle , construct midpoints  ,  and . Construct medians,  , and . Construct midpoints

and . Construct lines segments, parallel to , through points G and H. By similar triangles, we have FB = AF, so BG = GD and EC = AC, so HC = DH. Therefore, , since D = M( ). Take  to be a transversal of the three parallel lines,  , and . By the lemma, parallel lines divide all transversals in the same proportion, so we see that . Thus is divided into 3 equal points and P lies two-thirds of the distance from B to E. Likewise,  and P is two-thirds the distance from C to F, so  is concurrent with  at point P. Without loss of generality, P is two-thirds the distance from A to D. and the three medians are concurrent at P.

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