proving an equality in euclidean geometry 
Let $ABCD$ be a cyclic quadrilateral, $O = AC \cap BD$.
Let $M, N, P, Q$ be the midpoints of
$AB,BC,CD$ and $DA$, respectively, and $X, Y , Z, T$ be the projections of $O$ on $AB, BC,
CD$ and $DA$, respectively.
Let $U = MP \cap Y T$ and $V = NQ \cap XZ$.
Prove that
$\frac{UO}{VO} = \frac{AB \times CD}{BC \times DA}$
From drawing, I noticed a few things that should be true, such as that the quadrilaterals $XNYZ; QTPM; QXMN; BPZT; QXZT; PMNY$ should be cyclic,
and that the segment pairs $(QN, ZX) (TY, PM)$ should both be perpendicular.
I also thought about using Ptolemy's theorem, but I don't think any of the things I listed above can help with finding the LHS of the equation, and I'm out of ideas.
Edit: I've made some progress thanks to @saulspatz's suggestion about UOV being collinear, as it made me realize that perhaps I can do something using the circle with diameter UV, as it should pass through both the intersections of QN and PM, and ZX and TY. I'm still not quite sure how to prove this though, and how it would be useful.
 A: 
Let the midpoints of $AC$ and $BD$ be $E$ and $F$ respectively. Now , $\triangle EQN \sim \triangle OZX$ because $\angle QEN=\angle ZOX$ and $\frac {QE}{EN}=\frac {CD}{AB}=\frac {ZO}{OX}$. From here, by showing the angle at which corresponding sides of those triangles are inclined , we get that $QN\perp ZX$ and following the same argument, we can get $PM\perp YT$.
Now, let the midpoints of $OD$ and $OA$ be $K$ and $L$ respectively. Now, showing $\triangle ZKQ \cong \triangle QLX$ gives $QZ=QX$ and thereafter $QN$ is the perpendicular bisector of $ZX$ and hence $V$ is the midpoint of $ZX$. Similarly, $\triangle TOY\sim \triangle MFP$ and $U$ is the midpoint of $YT$.
Let $G$ be the midpoint of $PM$ and $QN$ ($MNPQ$ is a parallelogram). Now, $EG=FG$ because $EPFM$ is a parallelogram as well.
Then , $\frac {UO}{FG}=\frac {OT}{FM}$ (Because $\triangle TOY\sim \triangle MFP$)
$\Rightarrow UO=FG\cdot \frac {2OT}{AD}$
Similarly, $VO=EG\cdot \frac {2OX}{AB}=FG\cdot \frac {2OX}{AB}$
Hence , $\frac {UO}{VO}=\frac {OT}{OX} \cdot \frac {AB}{AD}$
Now, $\triangle TOX\sim \triangle DCB$(through angle chasing) and thereafter $\frac {OT}{OX}=\frac {CD}{BC}$.
Hence , $\frac {UO}{VO}=\frac {OT}{OX} \cdot \frac {AB}{AD}=\boxed {\frac {AB\cdot CD}{BC\cdot AD}}$
