Prove that the triangle formed by the midpoints between the vertices and the Nagel Point is tangent to the Spieker circle Consider triangle $ABC$ with midpoints of $BC, AC$ and $AB$ as $D, E$ and $F$ respectively, Spieker centre $S$ and Nagel point $N$. Let the midpoints of $AN, BN$ and $CN$ be $X, Y$ and $Z$ respectively. Show that the incircle of $DEF$ (the Spieker circle) is also the incircle of $XYZ$. 
I can prove that $XZ$ is parallel to $DF$. I think that the simplest proof would be to show that $XYZ$ and $DEF$ are congruent and homothetic centred at $S$, but $I$ cannot see how to do this.
Any solutions would be appreciated.
 A: Lemma 1. Let $I$ be the incenter of triangle $ABC$. Let the angle bisector $CI$ intersect $AB$ at point $L$ and let $T$ be the point of tangency of the incircle of triangle $ABC$ with the edge $AB$. Furthermore, assume that $P$ is the intersection point of $CT$ and the mid-segment $DE$. Then the incenter $I$ lies on the segment $FP$.
Proof: If for example one denotes by $BC = a, \, CA = b, \, AB = c$ one can calculate the lengths of the segments $$AF = \frac{c}{2}, \,\, \,  AT = \frac{b+c - a}{2} \,\,\, \text{ and } \,\,\, AL = \frac{bc}{a+b}$$ the length of $AL$ being calculated from the angle bisector theorem
$$\frac{AL}{BL} = \frac{CA}{BC} = \frac{a}{b}$$ Without loss of generality, assume that $a>b$. Then $$TF = AF - AT = \frac{a-b}{2} \,\, \text{ and } \,\, FL = AF - AL = \frac{(a-b)c}{2(a+b)}$$ Hence
$$\frac{TF}{FL} = \frac{a+b}{c}$$ Next, using the angle bisector theorem for $AI$ in triangle $ALC$, calculate the ratio $$\frac{LI}{IC} = \frac{AL}{AC} =  \frac{c}{a+b}$$  Furthermore, observe that since $DE$ is a mid-segment of $ABC$, the point $P = CT \cap DE$ is the midpoint of $CT$, so $CP = PT$. Finally, calculate the ratio
$$\frac{TF}{FL} \cdot \frac{LI}{IC}  \cdot \frac{CP}{PT} = \frac{TF}{FL} \cdot \frac{LI}{IC}  \cdot 1 = \frac{a+b}{c} \cdot \frac{c}{a+b} = 1$$   which by Menelaus' theorem implies that the points $F, \, I$ and $P$ are collinear, i.e. $I$ lies on $FP$. 
Lemma 2. The incenter $I$ of triangle $ABC$ is the Nagel point of the triangle $DEF$.
Proof: By doubling homothety from vertex $C$, triangle $EDC$ is mapped to $ABC$ so point $P$ is mapped to $T$. However, point $T$ is the touching point of the incircle of $ABC$ with $AB$, which means that $P$ is the touching point of the incircle of $EDC$ with $DE$. However, triangle $DEF$ is congruent to $CED$ and positioned in such a way that segment $FP$ is the perimeter bisecting segment of $DEF$ from vertex $F$. By, Lemma 1, point $I$ lies on $FP$. Analogously, by applying Lemma 1 to the other two perimeter bisecting segments of $DEF$ from vertices $D$ and $E$, one sees that $I$ lies on both of them, meaning that all three parameter bisecting segments of $DEF$ meet in $I$, which is exactly the definition of Nagel point.  
Let $G$ be the centroid of $ABC$ (and thus, the centroid of $DEF$ too). Then,  after performing the homothety that maps $DEF$ to $ABC$ with center $G$, the Nagel point $I$ of $DEF$ (Lemma 2) is mapped to the Nagel point $N$ of $ABC$. Therefore, $G$ lies on $IN$ and $\frac{IG}{GN} = \frac{1}{2}$. On the other hand, by the same homothety, the incenter $S$ of $DEF$ is mapped to the incenter $I$ of $ABC$ meaning that $G$ lies on $IS$ and $\frac{SG}{GI} = \frac{1}{2}$. Thus, said in different way, the point $N$ lies on $IG$ as well as the point $S$ also lies on $IG$. Hence, $S$ lies on $IN$ and $IS = SN$.
Concluding the proof of the problem. Performing homothety with center $N$ and coefficient $\frac{1}{2}$, one sees that the triangle $ABC$ is mapped to the triangle $XYZ$. Hence point $I$ is mapped to the midpoint of $IN$, which is $S$, as proven above. By the properties of homotheties, the incircle of $ABC$, whose center is $I$, is mapped to the incircle to $XYZ$, whcih is a circle with center $S$ and half the radius. However, there is only one such circle and that circle is the incircle of $DEF$ (the Spiker circle).  Hence, the incircle of $DEF$ is also the incircle of $XYZ$.   
