Convex quadrilateral; Area and parallel sides Consider a convex quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at $O$. 
Prove that the area of triangle $AOB$ = the area of the triangle $COD$ if an only if $AD$ and $BC$ are parallel.
Where should I start with this problem?
Thanks!
 A: Given the condition:
$$S_{AOB}=\frac12 AO\cdot BO\cdot \sin{\measuredangle AOB}\\
S_{COD}=\frac12 CO\cdot DO\cdot \sin{\measuredangle COD}\\
S_{AOB}=S_{COD} \Rightarrow \frac{AO}{OC}=\frac{DO}{BO}.$$
The triangles $AOD$ and $BOC$ are similar according to the property: 

If two triangles have two pairs of sides in the same ratio and the included angles are also equal, then the triangles are similar.

It implies $\measuredangle ADB=\measuredangle CBD$ and $\measuredangle CAD=\measuredangle ACB$. Hence $AD||BC$.
A: Triangles ABD and ACD have the same area having the same base and perpendicular height. Subtract the common triangle AOD from both and there is your answer. 
A: Good to have a diagram.

Now, if $AD || BC$ then $w = z$ and $x = y$, so we get that triangles $AOD$ and $BOC$ are similar. Hence, we get $\frac{OB}{OD} = \frac{OC}{OA}$, and therefore $OA \cdot  OB = OC \cdot OD$. We know that $\angle AOB = \angle COD$, hence from the formula for area,
$$
\mbox{Area}(AOB) = \frac 12 OA \cdot OB \cdot \sin \angle AOB = \frac 12 OC \cdot OD \cdot \sin \angle COD = \mbox{Area} (COD) 
$$ 
Conversely, if the areas are the same, then all steps reverse up till $OA \cdot OB = OC \cdot OD$. Let  $k = \frac{OD}{OB} = \frac{OA}{OC}$. Then, we see that:
$$ 
AD^2 = AO^2 + OD^2 - 2AO \cdot OD \cdot \cos \angle AOD \\ = k^2(OC^2 + OB^2 + 2 OC \cdot OB \cdot \cos \angle COB) = k^2 BC^2 \\ \implies \frac{AD}{BC} = k
$$
Hence, the triangles $AOD$ and $BOC$ are similar, hence have the same angles, and from the way the sides map, we see that $w = z$ and $x = y$, and hence $AD || BC$.
I feel I am doing something too complicated, so I would still like suggestions.
