Prove three points collinear A circle with center $O$ is inscribed in the convex quadrilateral $ABCD$. If $M$ and $N$ are the midpoints of the diagonals $AC$ and $BD$, prove that points $O$, $M$, and $N$ are collinear.
 A: This answer gives references, but only a sketch for the proof due to copyright concerns.
http://en.wikipedia.org/wiki/Tangential_quadrilateral#Collinear_points states this fact. It references T. Andreescu and B. Enescu, Mathematical Olympiad Treasures, p.42 for a proof. It also states that the line spanned by the diagonal midpoints $M$ and $N$ is called the Newton line. In the article about that, there is another reference with a proof, this time D. Djukić et al., The IMO Compendium. There is also a reference to Léon Anne's Theorem which is a generalization of this, based on areas. So if you had that theorem, proving your fact would be simple. Of course, the proof for that is buried in yet another book, R. Honsberger, More Mathematical Morsels.
Looking at Mathematical Olympiad Treasures p. 43 Problem 2.20, I read exactly your problem statement, including the comma after $M$. So it seems to me that you are dealing with past Mathematical Olympiad problem statements. It would have been nice if you had included that bit of context information in your question. The proof given in that book starts with the edge length equation for tangential quadrilaterals, then turns it into an equation for triangle areas with $O$ as third point. It obtains similar area equations for the other edges, and ends up observing that all points $X$ with $\operatorname{area}(XAB)+\operatorname{area}(XCD)=\tfrac12\operatorname{area}(ABCD)$ lie on a line, which is the condition already seen in Léon Anne's Theoreme, and which was proven in problem 2.19 building on problems before that.
I'm sorry that this may still be rather obscure, but I guess a more detailed reproduction of that proof would likely violate their copyright. So either treat the above as hints on which you can work, or follow this link to what appears to be a feely available digital version of this book, to have a closer look at the full proof.
Note that I found the online version of the book by pasting your problem statement to Google, enclosed in double quotes. Might be a useful technique for finding solutions to similar problems in the future.
A: Another approach would be brute force algebra. Start with the incircle, and place its center at the origin. Without loss of generality you can choose your unit of length such that its radius will be one. Then choose four angles $\alpha$, $\beta$, $\gamma$ and $\delta$ to denote the locations where the tangents touch the circle. From these you can compute the coordinates of the corners as
\begin{align*}
A&=
\frac1{ -\sin\left(\alpha\right) \cos\left(\beta\right) +
\sin\left(\beta\right) \cos\left(\alpha\right) } \left(\begin{array}{r}
-\sin\left(\alpha\right) + \sin\left(\beta\right) \\
\cos\left(\alpha\right) - \cos\left(\beta\right)
\end{array}\right)
\\
B&=
\frac1{ -\sin\left(\beta\right) \cos\left(\gamma\right) +
\sin\left(\gamma\right) \cos\left(\beta\right) } \left(\begin{array}{r}
-\sin\left(\beta\right) + \sin\left(\gamma\right) \\
\cos\left(\beta\right) - \cos\left(\gamma\right)
\end{array}\right)
\\
C&=
\frac1{ \sin\left(\delta\right) \cos\left(\gamma\right) -
\sin\left(\gamma\right) \cos\left(\delta\right) } \left(\begin{array}{r}
\sin\left(\delta\right) - \sin\left(\gamma\right) \\
-\cos\left(\delta\right) + \cos\left(\gamma\right)
\end{array}\right)
\\
D&=
\frac1{ \sin\left(\alpha\right) \cos\left(\delta\right) -
\sin\left(\delta\right) \cos\left(\alpha\right) } \left(\begin{array}{r}
\sin\left(\alpha\right) - \sin\left(\delta\right) \\
-\cos\left(\alpha\right) + \cos\left(\delta\right)
\end{array}\right)\end{align*}
You can then compute the midpoints, and check for the condition that these lie on a line with the center by evaluating the determinant
$$\begin{vmatrix}
A_x+C_x & B_x+D_x & 0 \\
A_y+C_y & B_y+D_y & 0 \\
2 & 2 & 1
\end{vmatrix}$$
Plugging this into a computer algebra system and massaging it in the right way will give you a nice zero as the result, confirming the fact that the three points lie on a line. I managed to do this using sage, using a quotient ring derived from a polynomial ring to handle all those trigonometric expressions and their relations.
sage: R.<ca,sa,cb,sb,cc,sc,cd,sd> = QQ[]
sage: tangents = Matrix(R,[[ca,sa,-1],[cb,sb,-1],
...                        [cc,sc,-1],[cd,sd,-1]]).rows()
sage: homogenous_corners = [
...     tangents[i].cross_product(tangents[(i+1)%4]) for i in range(4)]
sage: corners = [i/i[2] for i in homogenous_corners]
sage: M = Matrix([corners[0]+corners[2],
...               corners[1]+corners[3],[0,0,1]])
sage: I = R.ideal(ca^2+sa^2-1,cb^2+sb^2-1,cc^2+sc^2-1,cd^2+sd^2-1)
sage: QR = R.quotient(I)
sage: QR(M.det().numerator())
0

