Question 388219
<pre>
First we construct a triangle ABC whose sides are the medians,
with AC = 6, BC = 8, and AB = 10

{{{drawing(400,400,-2,12,-4,10,
locate(0,0,A), locate(10,0,B), locate(3.6,5.4,C),
red(line(0,0,10,0)), line(0,0,6cos(.927295218),6sin(.927295218)),

green(line(10,0,6cos(.927295218),6sin(.927295218)))  )}}}

Triangle ABC happens to be a right triangle, since 6² + 8² = 10², so

angle CBA has measure arccos({{{6/10}}}) = arccos(.6) = arcsin(8/10)=arcsin(.8).

We will let AB be an actual median of the final triangle we are 
going to create.

We know that the three medians intersect at a point which divides
each median into two parts such that the shorter part is {{{1/3}}} 
of that median, and, of course, the longer part is {{{2/3}}} of that
median.   

So we locate point D such that AD is {{{1/3}}} of AB and of course,
BD is {{{2/3}}} of AB.  So AD = {{{10/3}}} and BD = {{{20/3}}}

Point D is where all three medians of our final triangle will intersect.

{{{drawing(400,400,-2,12,-4,10,
locate(0,0,A), locate(10,0,B), locate(3.6,5.4,C),
red(line(0,0,10,0)), line(0,0,6cos(.927295218),6sin(.927295218)),
circle(10/3,0,.1), locate(10/3,0,D),
green(line(10,0,6cos(.927295218),6sin(.927295218)))  )}}}

Next we must draw a median-to-be EF of the final triangle through D parallel
and equal to AC such that point D divides median-to be EF such that DF is
{{{1/3}}} of median-to be EF and DE is {{{2/3}}} of median-to-be EF. Since
EF is 6, we will draw DF 2 units long and DE 4 units long:

{{{drawing(400,400,-2,12,-4,10,
locate(0,0,A), locate(10,0,B), locate(3.6,5.4,C),
red(line(0,0,10,0)), line(0,0,6cos(.927295218),6sin(.927295218)),
circle(10/3,0,.1), locate(10/3,0,D),
green(line(10,0,6cos(.927295218),6sin(.927295218))),
line(10/3,0,10/3+2cos(.927295218),2sin(.927295218)), locate(4.5,2.2,F),

line(10/3,0,10/3-4cos(.927295218),-4sin(.927295218)), locate(.8,-3.2,E)
  )}}}

Next we will draw line segment EG through A so that AE = AG. 
AG will be a side of the final triangle, and A will be the midpoint
of side EG of the final triangle.


{{{drawing(400,400,-2,12,-4,10,
locate(-.4,0,A), locate(10,0,B), locate(3.6,5.4,C),
red(line(0,0,10,0)), line(0,0,6cos(.927295218),6sin(.927295218)),
circle(10/3,0,.1), locate(10/3,0,D),
green(line(10,0,6cos(.927295218),6sin(.927295218))),
line(10/3,0,10/3+2cos(.927295218),2sin(.927295218)), locate(4.5,2.2,F),

line(10/3,0,10/3-4cos(.927295218),-4sin(.927295218)), locate(.8,-3.2,E),
line(0,0,(10/3)cos(1.854590436),(10/3)sin(1.854590436)), locate(-1,3.9,G),

line(0,0,10/3-4cos(.927295218),-4sin(.927295218)) 


  )}}}

Now we can calculate AE from triangle ADE by the law of cosines
since we have Side-Angle-Side  We know that 

side AD = {{{10/3}}}, 

angle ADE = angle EAB because they are alternate interior angles formed
by transversal AB cutting parallel lines AC and EF.

So angle ADE = arccos(.6)

side DE = 4

{{{AE^2=AD^2+DE^2-2AD*DE*cos(ADE)}}}

{{{AE^2=(10/3)^2+4^2-2(10/3)(4)cos(arccos(.6))}}}

{{{AE^2=(100/9)+16-(80/3)(.6)}}}

{{{AE^2=(100/9)+16-16}}}

{{{AE^2=100/9}}}

{{{AE = sqrt(100/9)}}}

{{{AE = 10/3}}}

And since A is the midpoint of EG,

side EG of the final triangle is twice that or 

side EG = {{{20/3}}}.  That's one of the sides of the final triangle.

Now if we have drawn everything right so far, points G, F, and B
should be colinear and GB should be a side of the final triangle,
with F at its midpoint.

{{{drawing(400,400,-2,12,-4,10,
locate(-.4,0,A), locate(10,0,B), locate(3.6,5.4,C),
red(line(0,0,10,0)), line(0,0,6cos(.927295218),6sin(.927295218)),
circle(10/3,0,.1), locate(10/3,0,D),
green(line(10,0,6cos(.927295218),6sin(.927295218))),
line(10/3,0,10/3+2cos(.927295218),2sin(.927295218)), locate(4.5,2.2,F),

line(10/3,0,10/3-4cos(.927295218),-4sin(.927295218)), locate(.8,-3.2,E),
line(0,0,(10/3)cos(1.854590436),(10/3)sin(1.854590436)), locate(-1,3.9,G),

line(0,0,10/3-4cos(.927295218),-4sin(.927295218)),

line(10,0,(10/3)cos(1.854590436),(10/3)sin(1.854590436))  


  )}}}
 
Next we must calculate FG.  But to do that we must calculate
angle E.  

We will use the law of sines in triangle ADE.

{{{AD/sin(E)=AE/sin(ADE)}}}

{{{sin(E)=AD*sin(ADE)/AE=(10/3)(sin(arcsin(.8)))/(10/3) = .8}}}

So angle E = arcsin(.8) = arccos(.6)

Now we can calculate FG by the law of cosines, since we have
side-angle-side.

EG = {{{20/3}}}, angle E = arccos(.6), EF = 6

{{{FG^2=EG^2+EF^2-2*EG*EF*cos(E)}}}
{{{FG^2=(20/3)^2+6^2-2(20/3)(6)cos(arccos(.6))}}}
{{{FG^2=400/9+36-80*.6}}}
{{{FG^2=400/9+36-48}}}
{{{FG^2=292/9}}}
{{{FG=sqrt(292)/3=sqrt(4*73)/3=(2sqrt(73))/3}}}

And since F is the midpoint of BG,

side BG of the final triangle is twice that or 

side BG = {{{4sqrt(73)/3}}}.  That's another side of the final triangle.

There is just one more side to find, and that is BE, which we will now
draw in:


{{{drawing(400,400,-2,12,-4,10,
locate(-.4,0,A), locate(10,0,B), locate(3.6,5.4,C),
red(line(0,0,10,0)), line(0,0,6cos(.927295218),6sin(.927295218)),
circle(10/3,0,.1), locate(10/3,0,D),
green(line(10,0,6cos(.927295218),6sin(.927295218))),
line(10/3,0,10/3+2cos(.927295218),2sin(.927295218)), locate(4.5,2.2,F),

line(10/3,0,10/3-4cos(.927295218),-4sin(.927295218)), locate(.8,-3.2,E),
line(0,0,(10/3)cos(1.854590436),(10/3)sin(1.854590436)), locate(-1,3.9,G),

line(0,0,10/3-4cos(.927295218),-4sin(.927295218)),

line(10,0,(10/3)cos(1.854590436),(10/3)sin(1.854590436)),  

line(10,0,10/3-4cos(.927295218),-4sin(.927295218)) 

  )}}}

We have now completed the final triangle BEG.  We only need side BE.

We can find it by the law of cosines since we have side-angle-side
in triangle BDE.

Side DE = 4, 

Angle BDE is supplementary to angle ADE and so its cosine is the
negative of the cosine of angle ADE, so angle BDE = arccos(-.6)

Side BD = {{{20/3}}}

{{{BE^2=DE^2+BD^2-DE*BD*cos(BDE)}}}
{{{BE^2=4^2+(20/3)^2-2*4*(20/3)cos(arccos(-.6))}}} 
{{{BE^2=16+(400/9)-(160/3)(-.6)}}}
{{{BE^2=832/9}}}
{{{BE=sqrt(832)/3=sqrt(64*13)/3=8sqrt(13)/3}}}

So we have found all three sides of triangle BEG.  
We didn't draw in the third median, but we don't need to.

Edwin</pre>