Constructing a Triangle with Three Medians
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
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.