Lines formed mid point of sides and cevians from opposite point are concurrent $AD, BE, CF$ are three concurrent lines in $\triangle ABC$ meeting opposite sides in $D, E, F$ respectively. Show that the joins of the midpoints of $BC, CA, AB$ to the midpoints of $AD, BE, CF$ are concurrent.  

(Should be done by Ceva's theorem, Menelaus theorem, Stewart's theorem) 
I tried by using trig form of Ceva's theorem and tried to do something similar to Cevian nests proof by connecting A'B'C' triangle but I failed. So please consider giving a hint or something and post the answer later on if I need it. 
Source:CTPCM
 A: Hints: 
(1) 
Pick 2 vectors e.g. 
$\overrightarrow{AB}$ and $\overrightarrow{AC}$
as a basis in the plane, and express all other vectors $\overrightarrow{AX}$ as linear combination of them (where X is any point on this drawing or whichever of the points you need to involve in your solution).    
This means, for any point X you should be able to find 2 numbers $k_1, k_2$ such that 
$\overrightarrow{AX} = k_1 \overrightarrow{AB} + k_2 \overrightarrow{AC}$ 
Then it's all a matter of picking good start (number) parameters and playing with the equations.  
Good parameters maybe are: 
AF:FB = a -> number   
AE:EC = b -> number 
Then BD:DC you can express in terms of a,b using Ceva.  
(2) Denote: 
$point\ S_1 = A'D' \cap B'E'$ 
$point\ S_2 = A'D' \cap C'F'$ 
Prove that $\overrightarrow{AS_1} = \overrightarrow{AS_2}$
If you do this, it would mean the points $S_1$ and $S_2$ coincide.  
My high school math is rusty but that (or similar) should be the general idea.   
(3)
Also, for arbitrary points in the plane ABCD, prove and use this lemma: 
$\overrightarrow{FE} = 1/2 (\overrightarrow{CA} + \overrightarrow{DB})$ 
(on the drawing below F and E are midpoints). 
Seems it might be useful for this problem.  

A: Look at the drawing here.    

What do we have?   
$AD,BE,CF$ - they intersect in a single/common point - point $O$
$A'$ - midpoint of $BC$
$B'$ - midpoint of $CA$
$C'$ - midpoint of $AB$ 
$D'$ - midpoint of $AD$
$E'$ - midpoint of $BE$
$F'$ - midpoint of $CF$ 
From Ceva's theorem for triangle $ABC$ we get: 
$$\frac{AF}{FB}\frac{BD}{DC}\frac{CE}{EA} = 1 \tag{1}$$ 
Now the trick is to realize that:
$$\frac{B'F'}{F'A'} = \frac{AF}{FB} \tag{2}$$
$$\frac{C'D'}{D'B'} = \frac{BD}{DC} \tag{3}$$
$$\frac{A'E'}{E'C'} = \frac{CE}{EA} \tag{4}$$ 
Why is this so?   
Because $B'C' || BC$   , $C'A' || CA$ and  $A'B' || AB$
so these relations follow from the Intercept theorem.     
Multiplying the last 3 equations and using $(1)$ we get:  
$$\frac{B'F'}{F'A'}\frac{C'D'}{D'B'}\frac{A'E'}{E'C'} = \frac{AF}{FB}\frac{BD}{DC}\frac{CE}{EA} = 1$$ 
Thus:  
$$\frac{B'F'}{F'A'}\frac{A'E'}{E'C'}\frac{C'D'}{D'B'} = 1 \tag{5}$$
Now using the inverse Ceva's theorem (for the triangle $A'B'C'$ and for the points $D', E', F'$), we can conclude from $(5)$ that the three lines  $A'D', B'E', C'F'$ intersect at a single/common point. This is what we had to prove hence the problem is solved.     
A: I would do this using Cartesian coordinates.  Given any triangle, we can set up a coordinate system with origin at one vertex and x-axis along one side.  In that coordinate system, the vertices of the triangle are at (0, 0), (a, 0), and (b, c) for some numbers, a, b, and c.  The midpoints of the three sides are (a/2, 0), (b/2, c/2), and ((a+b)/2, c/2).
The medians are given by y= ((a+b)/c)x, y= (2c/(2b-a))x+ 2c/(a-2b), and y= (ac/(4b-2a))x+ ac/(2a-4b).   Show that the solutions to any two of those equation are the same.
