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Fun problem on Ceva's theorem


Given is $\Delta ABC$. The points $D \in BC$, $E \in CA$, $F \in AB$ are such that the lines $AD,BE,CF$ are concurrent. 

$A'$ - midpoint of $BC$   
$B'$ - midpoint of $CA$   
$C'$ - midpoint of $AB$   
$D'$ - midpoint of $AD$  
$E'$ - midpoint of $BE$  
$F'$ - midpoint of $CF$  

Prove that the lines 

$A'D', B'E', C'F'$ are also concurrent (i.e. that they pass through a common point).


It is crucial to come up with a nice realistic drawing here.

Also, it is important to realize that:
a) $A', F', B'$ are on one line
b) $B', D', C'$ are on one line
c) $C', E', A'$ are on one line

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:

$$\frac{B'F'}{F'A'}\frac{C'D'}{D'B'}\frac{A'E'}{E'C'} = \frac{AF}{FB}\frac{BD}{DC}\frac{CE}{EA} = 1$$ 


$$\frac{B'F'}{F'A'}\frac{A'E'}{E'C'}\frac{C'D'}{D'B'} = 1 \tag{5}$$

Now using the converse 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.     

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