Newton's Theorem for tangential quadrilaterals is this:
The center of the circle inscribed into a quadrilateral lies on the line joining the midpoints of the latter's diagonals.
For more information, see Cut-the-Knot's entry.
Is there any synthetic proof which doesn't use Anne's theorem, trigonometry, or complex numbers? I've just found out a simple proof.
Here's my proof:
Lemma: Given $\triangle ABC$ with incenter $I$, let incircle $\bigcirc I$ touch $BC$ at $D$. Let $M$, $N$ be the midpoints of $BC$, $AD$. Then $\overline{M,I,N}$.
Proof: Let $DD^\prime$ be the diameter of $\bigcirc I$. A line through $D^\prime$ and parallel to $BC$ intersect $AB$, $AC$ at $E$, $F$. Let $AD^\prime$ meet $BC$ at $H$.
It's well-known that there is a homothety at $A$ which maps $\triangle AEF \leftrightarrow \triangle ABC$ and maps the excircle of $\triangle AEF$ (it's $\bigcirc I$, as well) into the excircle of $\triangle ABC$. Then $H$ is the tangent point of the excircle of $\triangle ABC$ with $BC$. So $M$ is the midpoint of $HD$. Therefore, $IM \parallel AH$. We also have $MN \parallel AH$, so done.
Now's the problem:
Given a tangential quadrilateral $\square SBCR$. Prove that $I$ is collinear with the midpoints of $CS$ and $BR$.
Let $BS$ intersect $CR$ at $A$, and let the incircle touch $BC$, $CR$, $RS$, $SB$ at $D$, $E$, $D^\prime$, $F$. Let $DD^\prime$ meet $EF$ at $Q$. It's well-known that $CS$, $BR$, $DD^\prime$, $EF$ are concurent at $Q$.
Let $M$, $N$, $P$, $X$, $Y$, $Z$, $T$ be the midpoints of $BC$, $CA$, $AB$, $BR$, $BE$, $CF$, $CS$.
From the lemma, we have $\overline{Y,I,N}$ and $\overline{Z,I,P}$. While $M$ is the midpoint of $BC$, we have: $\overline{M,Y,X,P}$ and $\overline{M,Z,T,N}$
It's well-known from the double ratio that $PZ$, $YE$, $XT$ are concurent if and only if $(M,Y,X,P)=(M,N,T,Z)$. We will prove this.
From Menelaus's theorem for $\triangle ABR$ we have $$\frac{\overline{CR}}{\overline{CA}}\cdot\frac{\overline{SA}} {\overline{SB}}\cdot\frac{\overline{QB}}{\overline{QR}}=1\tag{1}$$
Now, $$\begin{align} (M,Y,X,P)=(M,N,T,Z) &\iff \frac{\overline{XM}}{\overline{XY}}:\frac{\overline{PM}}{\overline{PY}}=\frac{\overline{TM}}{\overline{TN}}:\frac{\overline{ZM}}{\overline{ZN}} \\[6pt] &\iff \frac{\overline{RC}}{\overline{RE}}:\frac{\overline{AC}}{\overline{AE}}=\frac{\overline{SB}}{\overline{SA}}:\frac{\overline{FB}}{\overline{FA}} \\[6pt] &\iff \frac{\overline{RC}}{\overline{AC}}\cdot\frac{\overline{BF}}{\overline{RE}}\cdot\frac{\overline{SA}}{\overline{SB}}=1 \\[6pt] &\iff \frac{\overline{BF}}{\overline{RE}}=\frac{\overline{QB}}{\overline{QR}} \quad \text{(using $(1)$)} \\[6pt] &\iff \frac{BD}{RD^\prime}=\frac{QB}{QR} \\[6pt] &\iff \frac{BD}{BQ}=\dfrac{RD^\prime}{RQ} \\[6pt] &\iff \frac{\sin{\angle BQD}}{\sin{\angle BDQ}}=\frac{\sin{\angle RQD^\prime}}{\sin{\angle RD^\prime Q}} \end{align}$$ and this is easy to check its truth.
Nice solution! We can skip the computations by noticing that \begin{align} (M, Y, X, P) = (M, Z, T, N) &\iff (C, E, R, A) = (B, F, S, A) \\ &\iff CS \cap EF \cap RB \neq\varnothing\end{align} which is certainly the case.
