Constructing a Triangle with Three Medians


Jane Yun



Median Concurrency Theorem


The three median of any triangle are concurrent; that is, if ABC 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 AA, and mark the centroid G, which the 1/3 of the distance from A to A.














Extend AA to D, so that GA = DA'. 









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















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









Then, these two circles intersect at point C,

















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



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 proof


Given: In ABC, 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 ABK, 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 ACK, 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 ABC and AG:GD = BG:GE = CG:GF = 2:1.




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