Find a locus of points Given a triangle $ABC$, $A'$ and $B'$ halves $BC$ and $AC$. We have a variable point on a line $AB$. Parallel to $AA'$ and $BB'$ through $P$ cuts $AC$ in $E$ and $BC$ in $F$. Now line $EF$ cuts $AA'$ in $M$ and $BB'$ in $N$. Lines $A'N$ and $B'M$ meet at $Q$. What is a locus of $Q$?

When playing in Geogebra I found it could be an ellipse which goes through the midpoints of a segments $AG$, $BG$ and $CG$ where $G$ is a gravity center. Also, this ellipse seem to touches the sides of triangle at their midpoints. But when I was trying to write down equations of intersection points I get a headeacke due to huge formulas I get. Does anyone have nice idea how to avoid this equations?   
 A: Since the construction uses only affine-independent properties (incidence, collinearity, parallelism, midpoints) we can choose a convenient form of $\triangle ABC$. If we can show that the locus is the Steiner Inellipse for that form, then the locus is the Inellipse for every triangle.

We'll position vertex $C$ at the origin. Treating $A$ and $B$ as position vectors, we find the midpoints $A^\prime$ and $B^\prime$ as ...
$$A^\prime = \frac12(B+C) = \frac12 B \qquad B^\prime = \frac12(A+C) = \frac12 A \tag{1}$$
Define
$$P := \frac12(A+B)+\frac{p}{2}(B-A) \tag{2}$$
(where $p$ affects displacement from the midpoint of $\overline{AB}$), from which we derive
$$E = \frac{1}{4}(3-p)A \qquad F = \frac{1}{4}(3+p)B \tag{3}$$
and
$$\begin{align}
M &= \frac13( 2E+\phantom{2}F) = \,\,\frac16(3 - p) A + \frac1{12} (3 + p) B \\[6pt]
N &= \frac13(\phantom{2}E+2F) = \frac1{12}(3 - p) A + \frac16 (3 + p) B
\end{align} \tag{4}$$
(Note: The $EF$-representation shows that $M$ and $N$ trisect $\overline{EF}$.) And, finally, that
$$Q = \frac{1}{6(p^2+3)}\;\left(\;(p - 3)^2 A + (p + 3 )^2 B \;\right) \tag{5}$$
Now, let's locate $A$ and $B$ so that $\triangle ABC$ is an equilateral triangle of inradius $r$ (and circumradius $2r$), with the $x$-axis as an axis of symmetry:
$$A,B = 2r\sqrt{3}\left(\cos\frac{\pi}6,\pm\sin\frac{\pi}{6}\right) \tag{6}$$
Then 
$$Q = \frac{r}{p^2+3} \left(p^2+9, -2 p \sqrt{3}\right) = 2r (1,0) + \frac{r}{p^2+3}\left(3-p^2, -2p\sqrt{3}\right) \tag{7}$$
and we see that
$$\left(Q_x-2r\right)^2+Q_y^2 = \frac{r^2}{\left(p^2+3\right)^2}\left((3-p^2)^2 + 12 p^2\right) = r^2 \tag{8}$$
That is, the locus of $Q$ for this particular $\triangle ABC$ is the incircle, which is, in fact, the Steiner Inellipse for that form. $\square$
A: Lemma: $EM =MN = NF$
Proof: Let $PE\cap AA' = \{D\}$.
Then since $AEP\sim AB'B$ we have $${PD\over DE} = {BG'\over GB'} = {2\over 1}$$ But $DEM\sim PEF$ so $${EM\over EF} = {PD\over DE} ={2\over 1} $$
By symmetry we have $${FM\over FE} = {2\over 1} \;\;\;\;\;\;\;\blacksquare{}$$ 

Now map $E\mapsto F$ is projective transformation from the line $AC$ to the line $BC$ which induces projective map $M\mapsto N$ the line $AA'$ to the line $BB'$. Now this one induces projective transformation $A'N \mapsto B'M$ from a pencil of lines through $A'$ to a pencil of lines through $B'$ and thus $Q\in A'N\cap B'M$ describe some line or a conic. 
Now let's take a look some special positions of $P$ and try to determine where $Q$ is. 
$\bullet $ If $P=A$ then $E=A$ and $F=A'$ so $N=G$ (lemma) and $M$ halves $AG$ (lema), so $Q$ also halves $AG$. 
$\bullet $ If $P=B$ then by symmetry $Q$ also halves $BG$. 
$\bullet $ If $P=C'$ then $Q=C'$.
$\bullet $ If $P$ is such that $F = C$ then $Q = B'$. 
$\bullet $ If $P$ is such that $E = C$ then $Q = A'$. 
So this conic goes through $A',B',C'$ and halves $AG$ and $BG$ so this is Steiner ellipse (remember that conic is uniqely determined with $5$ noncolinear points). 
Thank you guys with usefull instructions.
