geometrytriangles
$AD$, $BE$, $CF$ are concurrent lines in $\triangle ABC$. Show that the lines through the midpoints of $BC$, $CA$, $AB$ respectively parallel to $AD$, $BE$, $CF$ are concurrent.
My attempt
I tried using Ceva's theorem but not getting how to use it. Please give me a hint.
I know this question has already been answered (for instance, see here), but I want hint related to Ceva's or Menelaus' theorem.
Yes, Ceva's a good choice here. Let's recall another version of Ceva's theorem
In the above picture, the three lines $XI, YH, ZG$ are concurrent if and only if $$\frac{YI}{IZ}\cdot \frac{ZH}{HX} \cdot\frac{XG}{GY} = 1.$$ Now, the ratios of the lengths are equal to the ratios of the areas of certain triangles, more specifically: $$\frac{YI}{IZ} = \frac{\triangle XYI}{\triangle XZI},\dots$$ here, we simply denote the triangle by its area. But $$\triangle XYI = \frac12 XY\cdot XI \cdot \sin (\angle YXI), $$ and similar for $\triangle XZI$. So $$\frac{YI}{IZ} = \frac{XY}{XZ}\cdot \frac{\sin\angle YXI}{\sin \angle ZXI}.$$ Taking cyclic product, we have the sine version of Ceva's theorem: That the three lines $XI, YH, ZG$ are concurrent if and only if the product $\frac{\sin\angle YXI}{\sin \angle ZXI} \cdot \frac{\sin\angle ZYH}{\sin \angle XYH} \cdot \frac{\sin\angle XZG}{\sin \angle YZG}$ of cyclic ratios of sines is equal to $1$.
Now that's what we will use in this problem:
So we want to show that $$\mathcal{F} = \frac{\sin\angle BAP}{\sin\angle PAC}\cdot \frac{\sin\angle ACR}{\sin\angle RCB} \cdot \frac{\sin\angle{CBQ}}{\sin\angle QBA} = 1.\tag{1}$$ Now $$\frac{\sin\angle BAP}{\sin\angle PAC} \cdot \frac{AF}{AE} =\frac{\triangle AFP}{\triangle AEP} = \frac{FP}{PE},$$ and so on. Taking the cyclic product of the above, we obtain $$\mathcal{F} \cdot \frac{AF}{AE}\cdot\frac{BD}{BF}\cdot \frac{CE}{CD} = \frac{FP}{PE}\cdot\frac{ER}{RD}\cdot\frac{DQ}{QF}.$$ And (1) follows from classic Ceva's theorem with the given concurrences.
The original triangle $ABC$ and its medial triangle $M_1 M_2 M_3$ are related via a dilation $\delta$ centered at the common centroid $G$. Any dilation preserves parallel lines and concurrency, so if $AD,BE,CF$ are concurrent so they are their images via $\delta$, i.e. the given lines through $M_1,M_2,M_3$.
Best Answer
Let $\Delta ABC$ be a triangle, denote by $M,N,P$ the mid points of the sides opposite to $A,B,C$. Let $D$ be a point on $BC$. The parallel to $AD$ through $M$ intersects $NP$ in a point denoted by $D'$.
Construct $E',F'$ in a similar way on the sides of $\Delta MNP$.
Because $MN$, $ND'$, $D'M$ are parallel to (respectively) $AB$, $BD$, $DA$, we obtain a triangle similarity, so $D'N:DB=MN:AB=1:2$. This gives $D'N:D'P=DB:DC$.
We finally apply Ceva (and the reciprocal) in the two triangles $\Delta ABC$, $\Delta MNP$: $$ \frac{D'N}{D'P}\cdot \frac{E'P}{E'M}\cdot \frac{F'M}{F'N} = \frac{DB}{DC}\cdot \frac{EC}{EA}\cdot \frac{FA}{FB} = -1\ . $$ $\square$
Later edit: Quick picture inserted: