Where is the homothety in the problem?

Question

I have to solve the following problem using homothety but I don't see where it is.

Given triangle $ABC$. $D$ is an arbitrary point inside the triangle. Points $M, E$ and $F$ are mid points of the sides $AB, AC$ and $BC$ respecitvely. Points $N, P$ and $Q$ are midpoints of $DM, DE$ and $DF$. Prove that the lines $AQ, BP$ and $CN$ intersect at a point.

Answer 1

A homothety with center $G$ (the point where your lines coincide) and factor $-4$ will map $\triangle QPN$ to $\triangle ABC$.

To see this, use $G$ as the center of a coordinate system, assume the homothety, and conclude the collinearity in $D$:

\begin{align*} Q &= -A/4 & P &= -B/4 & N &= -C/4 \\ M &= (A+B)/2 & E &= (A+C)/2 & F &= (B+C)/2 \\ D &= N + (N - M) & D &= P + (P-E) & D &= Q + (Q - F) \end{align*}

All three descriptions of $D$ lead to

$$ D = -(A+B+C)/2 $$

so the assumed homothety is consistent with the construction starting from $D$.

Answer 2

Here $PNQ$ is Homothetic to $EMF$ with center D and ratio $=2$

And $EMF$ is Homothetic to $BCA$ with center centroid of the triangle $BCA$ and ratio $=-2$

$\implies PNQ$ is Homothetic to $BCA$ with ratio $=2\times(-2)=-4$ (As it will be composition of above two homotheties). It’s center will be point of concurrency of $QA, NC, PB.$

Q.E.D.