HELP On the sides of the triangle $ABC$ we consider the points $M,N,P$ so that $\frac{MB}{MC}=\frac{NC}{NA}=\frac{PA}{ PB}=k$, where $k\neq1$.

Question

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!

Answer 1

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 $AN$).

Answer 2

Here is a solution using vector geometry.

Because $MB/MC = k$, we have $$ \overrightarrow{B}-\overrightarrow{M} = k(\overrightarrow{M}-\overrightarrow{C}), \quad \mbox{so} \quad (k+1) \overrightarrow{M} = \overrightarrow{B} + k \overrightarrow{C}. $$ Similarly, we have $$ (k+1)\overrightarrow{P} = \overrightarrow{A} + k \overrightarrow{B} \quad \mbox{and} \quad (k+1)\overrightarrow{N} = \overrightarrow{C} + k \overrightarrow{A}. $$ Because $BDCP$ is a parallelogram by construction, $$ \overrightarrow{D} = \overrightarrow{P} - \overrightarrow{B} - \overrightarrow{C}. $$ We now calculate $$ \begin{aligned} \overrightarrow{A} - \overrightarrow{M} - \overrightarrow{N} &= \overrightarrow{A} - \frac{1}{k+1} \overrightarrow{B} - \frac{k}{k+1} \overrightarrow{C}-\frac{1}{k+1}\overrightarrow{C} - \frac{k}{k+1} \overrightarrow{A} \\ &= \frac{1}{k+1} \overrightarrow{A} + \frac{k}{k+1} \overrightarrow{B} - \overrightarrow{B} - \overrightarrow{C} \\ &= \overrightarrow{P} - \overrightarrow{B} - \overrightarrow{C} = \overrightarrow{D}, \end{aligned} $$ so $AMDN$ is a parallelogram as well.