$AD, BE, CF$ 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
Best Answer
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.