PROBLEM
On the sides $BC,CA,AB$ of the triangle $ABC$ we consider the points $M,N,P$ so that $\frac{MB}{MC}=\frac{NC}{NA}=\frac{PA}{ PB}=k$, where $k\neq1$. The parallel through $B$ to $PC$ and the parallel through $C$ to $AB$ intersect in $D$. Prove that $AMDN$ is a parallelogram.
WHAT I THOUGHT OF
the drawing
So as you can see, I came with the idea of letting a point $E$ that equals $AP$. $E$ is collinear with $D,C$.
We can easily demonstrate that $PCDB$ is a parallelogram.
After this we can show that triangles $PAC$ and $DEB$ are congruent.
Then, we arrive to the conclusion that $BAC,CEB$ are also congruent.
I think these will help us demonstrate that $AN=DM, AM=ND$ by some congruences, but I cant really figure it out. Hope one of you can help me! Thank you!
Best Answer
As you already said, $PCDB$ is a parallelogram, so $CD = BP$, and $D\hat{C}M = P\hat{B}M$. But then, triangles $\triangle CDM$ and $\triangle BAC$ are similar, since $\dfrac{CD}{AB} = \dfrac{CM}{BC}$ and $D\hat{C}M = P\hat{B}M$. The ratio between the third side will be the same, and so $\dfrac{DM}{AC} = \dfrac{CD}{AB} = \dfrac{CM}{BC}$. But note that $\dfrac{AN}{AC} = \dfrac{CM}{BC}$, so $DM = AN$. Also, since $\triangle BAC \sim \triangle CDM$ as proved before, we have $C\hat{M}D = A\hat{C}B$. But this means lines $DM$ and $AC$ are parallel. Therefore, $AMDN$ is a parallelogram, because one pair of opposite sides is parallel and equal in length (sides $DM$ and $AN$).