Use power of a point etc and it comes out nicely. Thats one approach. Bomb. Mettle strikes agains!

Label the quadrilateral internal angles $\alpha = \angle A, \beta = \angle B, \gamma = \angle C, \delta = \angle D$. Let $E, F, G, H$ be the tangency points of the sides $AB, BC, CD, DA$ with the incircle $(O)$, respectively. Let the opposite sides $AB, CD$ and $BC, DA$ intersect at points $X, Y$, respectively. WLOG, assume that the incenter $O$ is inside of the triangles $\triangle ABX, \triangle BCY$. (1) Let $K, L, M, N$ the midpoints of the sides $AB, BC, CD, DA$ and assume that the intersection of the midlines $KM, LN$ is identical with the quadrilateral incenter $O$. The circle $(O)$ is the common incircle of the triangles $\triangle ABX, \triangle BCY$. The bisector $XO$ of the angle $\xi = \angle BXA$ is perpendicular to the chord $FH$ and intersects it at its midpoint $X'$. Let a line through the incenter $O$ parallel to the chord $FH$ intersect the lines $BC \equiv BX$ and $DA \equiv XA$ at points $L', N'$. From similarity of the isosceles triangles $\triangle FXH \sim \triangle L'XN'$, we have $OL' = ON'$. Since the quadrilateral $KLMN$ is a parallelogram, its diagonals $KM, LN$ intersecting at the incenter $O$ cut each other in half, i.e., $OL = ON$ as well. As this is possible for one line only, the points $L \equiv L', N \equiv N'$ are identical and the lines $LN \parallel FH$ are parallel. The triangle $\triangle EFH$ is the contact triangle of the triangle $\triangle ABX$, hence, its internal angles are equal to $\angle HEF = \frac{\alpha + \beta}{2},\ \ \angle EFH = \frac{\beta + \xi}{2},\ \ \angle FHE = \frac{\xi + \alpha}{2}$ It follows that the angles $\angle BLN = \angle ANL$ are equal: $\angle BLN = \angle BFH = \angle BFE + \angle EFH = 90^o - \frac{\beta}{2} + \frac{\beta + \xi}{2} = 90^o + \frac \xi 2$ $\angle ANL = \angle AHF = \angle AHE + \angle FHE = 90^o - \frac{\alpha}{2} + \frac{\xi + \alpha}{2} = 90^o + \frac \xi 2$ The angles $\angle OBL = \angle AON$ are also equal: $\angle OBL = \frac \beta 2,\ \ \angle AON = 180^o - \left(\frac \alpha 2 + 90^o + \frac \xi 2\right) = 90^o - \frac{\alpha + \xi}{2} = \frac \beta 2$ Thus the triangles $\triangle OBL \sim \triangle AON$ are similar. In exactly the same way, it can be shown that the triangles $\triangle OCL \sim \triangle DON$ are also similar. Consequently, $\frac{OA}{OB} = \frac{ON}{LB} = \frac{NA}{OL},\ \ \frac{OD}{OC} = \frac{ON}{LC} = \frac{ND}{ON}$ Since $L, N$ are the midpoints of the quadrilateral sides $BC, DA$ and $O$ is the midpoint of the parallelogram diagonal $LN$, i.e., $LC = LB, NA = ND$, $ON = OL$, it follows that $\frac{OA}{OB} = \frac{OD}{OC},\ \ OA \cdot OC = OB \cdot OD$ (2) Conversely, assume that $OA \cdot OC = OB \cdot OD$ and let a line through the incenter $O$ parallel to the chord $FH$ intersect the lines $BC \equiv BX$ and $DA \equiv XA$ at points $L, N$. Obviously, $OL = ON$ (see the argument for the case (1)). In exactly the same way as in the case (1), we arrive to the relations $\frac{OA}{OB} = \frac{ON}{LB} = \frac{NA}{OL},\ \ \frac{OD}{OC} = \frac{ON}{LC}= \frac{ND}{OL}$ Since $\frac{OA}{OB} = \frac{OD}{OC}$, it follows that $\frac{ON}{LB} = \frac{ON}{LC}\ \Rightarrow\ LB = LC$ $\frac{NA}{OL} = \frac{ND}{OL}\ \Rightarrow\ NA = ND$ Thus the points $L, N$ are midpoints of the sides $BC, DA$ and $O$ is the midpoint of the segment $LN$. It follows that the diagonals of the parallelogram $KLMN$, where $K, M$ are midpoints of the sides $AB, CD$, intersect at the incenter $O$.

One can use complex numbers in this problem. Let $x, y, z, w$ the points where the incircle of $ABCD$ touches $AB$, $BC$, $CD$ and $DA$, respectively. Suppose that this circle is the unit circle with center on the origin. Thus $A = {2xw\over x+w}$, $B = {2xy\over x + y}$, $C = {2yz\over y + z}$ and $D = {2zw\over z+w}$. We must prove that $A + B + C + D = 0 \iff |OA|\cdot|OC| = |OB|\cdot |OD|$, that is, ${2xw\over x+w} + {2xy\over x + y} + {2yz\over y + z} + {2zw\over z+w} = 0 (1) \iff |A\cdot C| = |B\cdot D| (2)$. We'll expand $(1)$ and $(2)$ and see what happens. Recall that $\overline x = 1/x$ and so on. $(1) \iff {xw(y+z) + yz(x+w)\over (y+z)(z+w)} + {xy(z+w) + zw(x+y)\over (y+z)(z+w)} = 0 \iff xyz + xyw + xzw + yzw = 0$ or $(y+z)(x+w) + (z+w)(x+y) = 0\iff \overline x + \overline y + \overline z + \overline w = 0$ or $(y+z)(x+w) + (z+w)(x+y) = 0\iff x + y + z + w = 0$ or $(y+z)(x+w) + (z+w)(x+y) = 0$. $x + y + z + w = 0$ means that the midpoint of the lines connecting $x, y$ and $z, w$ is the center of the circle, which menas that $ABCD$ has parallel sides. So $(1)$ is equivalent to $(y+z)(x+w) + (z+w)(x+y) = 0$. $(2) \iff \left|{2xw\over x+w}\cdot {2yz\over y + z}\right| = \left|{2xy\over x + y}\cdot {2zw\over z+w}\right| \iff |(x+w)(y+z)| = |(x+y)(z+w)| \iff (x+w)(y+z)(\overline x + \overline w)(\overline y + \overline z) = (x+y)(z+w)(\overline x + \overline y)(\overline z + \overline w) \iff \bigl((x+w)(y+z)\bigr)^2 = \bigl((x+y)(z+w)\bigr)^2 \iff (x+w)(y+z) = (x+y)(z+w)$ or $(x+w)(y+z) + (x+y)(z+w) = 0$. But $(x+w)(y+z) = (x+y)(z+w) \iff xy + xz + yw + zw = xz + xw + yz + yw \iff (x-z)(y-w) = 0\iff x = z$ or $y = w$, contradiction. So $(2)$ is also equivalent to $(y+z)(x+w) + (z+w)(x+y) = 0$ and we're done.

I have just obtained a different solution of this problem. It is presented in the note "Circumscribed quadrilaterals revisited" on my website (you only need the very beginning of the note, i. e. the notations, the statement and proof of Theorem 11, and the statement and proof of Theorem 13). Darij

Let $ M$, $ N$, $ P$, $ Q$ the tangency points of $ (O)$ with the sides $ AB$, $ BC$, $ CD$, $ DA$, respectively. Denote $ x = AM = AQ$, $ y = BM = BN$, $ z = CN = CP$, $ t = DP = DQ$. We have $ (y + t)\left(\overrightarrow{OA} + \overrightarrow{OC}\right) + (z + x)\left(\overrightarrow{OB} + \overrightarrow{OD}\right) = \overrightarrow{0}$ Hence, $ \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = \overrightarrow{0}$ if and only if $ (y + z - t - x)\left(\overrightarrow{OA} + \overrightarrow{OC}\right) = 0$, i.e., $ y + t = z + x$ Now, notice that $ x = R\cdot\cot\frac {A}{2}$, and similarly for $ y$, $ z$, $ t$. After some of calculation, we get $ O$ is the centroid of $ ABCD$ if and only if $ \sin\frac {B}{2}\sin\frac {D}{2} = \sin\frac {A}{2}\sin\frac {C}{2}$, i.e., $ OA\cdot OC = OB\cdot OD$

This is one of the three problems this year copied from Russia (final round) 2005.

Problem 1 set 6 of India Postal Coaching 2011 is similar to this problem http://www.artofproblemsolving.com/Forum/resources.php?c=78&cid=50&year=2011&sid=b64ff370b3c1016346aaf4dc37628878

Probably what April did, but in a bit more detail. We will find an equivalent condition for both of them. Let $\alpha, \beta, \gamma, \delta$ denote $\frac12 \angle A, \frac12 \angle B, \frac12 \angle C,$ and $\frac12 \angle D$ respectively. Then, suppose WLOG that rays $AD$ and $BC$ meet, say at $P$. Let $M_1, M_2$ denote the midpoints of $AD, BC$ respectively and let $T_1, T_2$ be the points where $AD, BC$ touch the circle. If $O$ is the barycenter, then $O$ is the midpoint of $M_1M_2$, and so we have by symmetry that $M_1T_1 = M_2T_2$. This implies that $AT_1 - DT_1 = BT_2 - CT_2,$ which is equivalent to $R(\cot \alpha - \cot \delta) = R(\cot \beta - \cot \gamma),$ where $R$ is the radius of the circle. This rearranges into $\cot \alpha + \cot \gamma = \cot \beta + \cot \delta$. Now, observe that $OA = \frac{R}{\sin \alpha}$, etc., and so $OA \cdot OC = OB \cdot OD$ is equivalent to $\sin \alpha \cdot \sin \gamma = \sin \beta \cdot \sin \delta.$ It therefore now suffices to show that when $\alpha + \beta + \gamma + \delta = 180$, we have that $\sin \alpha \cdot \sin \gamma = \sin \beta \cdot \sin \delta \Leftrightarrow \cot \alpha + \cot \gamma = \cot \beta + \cot \delta.$ Indeed, observe that $\sin \alpha \cdot \sin \gamma (\cot \alpha + \cot \gamma) = \cos \alpha \cdot \sin \gamma + \sin \alpha \cdot \cos \gamma = \sin (\alpha + \gamma) = \sin (\beta + \delta) = sin \beta \cdot \cos \delta + \cos \beta \cdot \sin \delta = \sin \beta \cdot \sin \delta (\cot \beta + \cot \delta).$ This clearly implies the desired conclusion. $\square$

Here’s (i think)a different approach. Lemma $1$: $ABCD$ is a quadrilateral then let $E,F,G,HI,J$ be the middles of $AB,BC,CD,DA,AC,BD$. Then the middles of $EG,FH,IJ$ are the same point. (For proof just compute the barycenter of ABCD in two different ways) Lemma 2: points $A,B,C$ are collinear(with the same sequence) so are $E,F,G$ and $H,I,J$. also $A,E,H$ and $C,G,J$ are respectively collinear. Also $\frac{AB}{BC}= \frac{EF}{FG}= \frac{HI}{IJ}$ and $\alpha= \frac{AE}{EH}= \frac{CG}{GJ}$. Then $B,F,I$ are collinear and $\alpha= \frac{BF}{FI}$. (for proof give useful weights to $A,C,J,H$ [not necessarily positive] then find their barycenter in two different ways) Lemma 3: in triangle $ABC$, point $D$ is on $BC$. Then $\frac{BD}{CD}=\frac{AB}{AC} \cdot \frac{\sin{\widehat{ ABD}}}{\sin{\widehat{ ACD}}}$ Back to the problem:(i used $I$ instead of $O$) Let $AD,BC$ intersect in $T$. Let line perpendicular to $IT$ Intersect $AD$,$BC$ in $E,F$. call the middle of $AC,BD$ points $N,M$. Note that the barycenter of $ABCD$ is in the middle of $MN$ and I is in the middle of $EF$. Simple angle chasing gives us $\widehat{AIE}= \widehat{IBC}$ and $\widehat{EID}= \widehat{ICB}$. So using lemma 3 on $AID$ and $E$ we get $\frac{AE}{DE}=\frac{AI}{DI} \cdot \frac{\sin{\widehat{AIE}}}{\sin {\widehat{DIE}}}=\frac{AI\cdot CI}{BI \cdot DI}$. Similarly we get $\frac{AE}{DE}= \frac{AI\cdot CI}{BI \cdot DI}=\frac{CF}{BF}$. now using lemma 2 on $A,E,D$-$N,I,M$-$C,F,B$ we get $N,I,M$ are collinear and $\frac{NI}{MI}= \frac{AI\cdot CI}{BI \cdot DI}$ using the fact that the barycenter is in the middle of $NM$ the conclusion is obvious.