the question
Consider the tetrahedron $ABCD$ and $M$ a point inside the triangle $BCD$. Parallels taken from $M$
to the edges $AB$, $AC$, $AD$ intersect the faces $(ACD)$, $(ABD)$, respectively, $(ABC)$ at the points $A', B',$
respectively, $ C'$. If
$(BCD) || (A' B 'C')$
, prove that $M$ is the centroid of the triangle
$BCD$.
my idea
the drawing
So we have $B'M|| AC, A'M||AB, C'M||AD$
As you can see, I intersected the plane $(A'B'C')$ with $AD,AC,AB$ in $X,Y,Z$
We can simply demonstrate that $Y,A',X$ are collinear Analogus, $X,B',Z$ and $Z,C',Y$ are collinear
We can demonstrate that because $(ZYX)||(BCD)$ and both planes are intersected by 2 parallel lines such as BZ and A'M , we get that BZ=A'M which makes BMA'Z a parallelogram
Analogus, XC'MD and YB'MC parallelograms.
I don't know what to do going forward! Hope one of you can help me! Thank you!
Best Answer
This answer uses $X,Y,Z$ that you defined.
Let $E$ be a point on $BC$ such that $ME\parallel CD$.
Let $F$ be a point on $CD$ such that $MF\parallel BD$.
Let $G$ be a point on $BD$ such that $MG\parallel BC$.
Then, we have $$AD:AX=BC:BE\tag1$$ (Proof : Let us consider the plane $\alpha$ on which $A,A',B$ and $M$ exist. Let $P$ be the intersection point of $\alpha$ with $CD$. Then, we have $PA:AA'=PB:MB$. We also have $PA:AA'=AD:AX$ and $PB:MB=BC:BE$. So, we get $AD:AX=BC:BE$.$\ \square$)
Similarly, we have $$AD:AX=CD:CF\tag2$$ $$AC:AY=DB:DG\tag3$$
Now, if $(BCD)\parallel (A'B'C')$, then we can say that $AD:AX=AC:AY$.
So, it follows from $(1)(2)(3)$ that $$BC:BE=CD:CF=DB:DG\tag4$$
Let $H$ be a point on $BD$ such that $MH\parallel CD$.
Let $I$ be a point on $BC$ such that $MI\parallel BD$.
Let $J$ be a point on $CD$ such that $MJ\parallel BC$.
We have $$CJ=FD\tag5$$
By Menelaus's theorem, we get $$\frac{CF}{FP}\times\frac{PM}{MB}\times\frac{BI}{IC}=1$$ Since $\frac{BI}{IC}=\frac{DF}{CF}$, we have $$\frac{PM}{MB}=\frac{FP}{DF}\tag6$$
By Menelaus's theorem, we get $$\frac{DJ}{JP}\times\frac{PM}{MB}\times\frac{BG}{GD}=1$$ Since $\frac{BG}{GD}=\frac{CJ}{JD}$, we have $$\frac{PM}{MB}=\frac{JP}{CJ}\tag7$$
It follows from $(5)(6)(7)$ that $$JP=FP\tag8$$
It follows from $(5)(8)$ that $P$ is the midpoint of $CD$.
Similarly, we can see that the intersection point of $CM$ with $BD$ is the midpoint of $BD$.
Therefore, we can say that $M$ is the centroid of the triangle $BCD$.
Added :
Since $A'M\parallel AB$, the line $AA'$ intersects the line $BM$. Let $Q$ be the intersection point of $AA'$ with $BM$. We see that $Q$ is on $\alpha$. We also see that $Q$ is both on the plane $ACD$ and on the plane $BCD$. So, we see that $Q$ is on the line $CD$. It follows that $P=Q$. So, $P$ is both on the line $AA'$ and on the line $BM$.
I replaced $K$ with $P$ since $K$ is nothing but $P$.
$PA:AA′=PB:MB$ since $\triangle{PAB}\sim\triangle{PA'M}$.
$PA:AA′=AD:AX$ since $\triangle{AA'X}\sim\triangle{APD}$.
$PB:MB=BC:BE$ since $\triangle{BME}\sim\triangle{BPC}$.