Constructing a Triangle with Three Medians

by

Jane Yun

   

 

Median Concurrency Theorem

 

The three median of any triangle are concurrent; that is, if ÆABC and D, E, & F are the midpoints of the sides opposite A, B, & C, respectively, then the segments AF, BD, & CE all intersect in a common point G. Moreover, AG = 2GD, BG = 2 GE, and CG = 2GF.

 

We know that medians trisect each other, and that trisect point is called centroid.  The centroid (G) of a triangle is the common intersection of the three median points.  The median of a triangle is the segment from a vertex to the midpoint of the opposite side.

 

 

We will explore and investigate how to construct a triangle with three segments that are medians.

 

Here are the give three segments that are medians.

 

 

 

 

 

Draw the median AAÕ, and mark the centroid G, which the 1/3 of the distance from AÕ to A.

 

 

 

 

 

 

 

 

 

 

 

 

 

Extend AAÕ to D, so that GAÕ = DA'. 

 

 

 

 

 

                                                      

 

 

Draw a circle with the center G and radius 2/3 of CCÕ.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Again, draw a circle with center D and radius 2/3 of BBÕ.

 

 

 

 

 

 

 

 

Then, these two circles intersect at point C,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

And we can locate B in any of several ways and draw the ÆABC.

 

 

In our finish construction, quadrilateral BGCD will be a parallelogram, where the segments GD & BC are the diagonals bisect each other.

 

 

 

 

Want to see my GSP and check my work!! Click here.

 

 

Also here is my proofÉ

 

Given: In ÆABC, the median BE & CF intersects at G.  AG is joined and produced to meet BC in D.

 

Prove that AD is the median, the segment BD = the segment DC, and the ratio of AG:GD, BG:GA,GE:CG is 2:1

 

 

 

Proof:

 

Produce AD to K such that AG=GK.  Join KB and KC.

 

First, we want to show that the segment AD is the median, so it divides the segment BD into BD = DC.

 

Then, in ÆABK, F is the midpoint of AB since CF is the median, the segment CF makes the point F is the midpoint (by definition of median).

 

G is the midpoint of AK by the construction.

 

Therefore, FG║BK or GC║BK (by the theorem on line joining the midpoints of any two sides of a triangle.)

 

Similarly, in ÆACK, GE║CK or GB║CK and GE = ½ CK

 

Since GB║CK & GE = ½ CK, BGCK is a parallelogram.

 

Since BGCK is a parallelogram, the segments BC and GK are diagonals.

And, the diagonals of a parallelogram bisect each other,

so the segment BD = DC.

 

Therefore, the segment AD is the median from A since the BD = DC and by the definition of median.

 

Next, I want to show that the ratio is the 2:1.

 

Since GK and BC are the diagonals of BGCK, GD = DK.

 

So it follows that GD = ½ GK and GD = ½ AG.

 

Therefore, the AG:GD = 2:1.

 

Similarly, we can prove that BG:GE = 2:1 and CG:GF = 2:1.

 

Therefore, the segment AD is the median of ÆABC and AG:GD = BG:GE = CG:GF = 2:1.

 

QED

 

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