Problem based on proving equal triangles in area In the figure, $ABCD$ is a parallelogram. If $O$ be any point on $BD$ then prove that $$\triangle OAB=\triangle OAD+\triangle OAC$$
My Attempt 
$$\triangle ADB=\triangle BDC$$
$$\triangle OAM=\triangle OMC$$
Now, please help me to move further..
 A: Hint:
You have proved that $$\triangle OAM=\triangle OMC$$
Now note that $\overline{DO}+\overline{OM}=\overline {MB}$ so that :
$$
\triangle OAD+\triangle OAM=\triangle AMB
$$
because the triangles have the same height, and
$$
\triangle AOB=\triangle AOM+\triangle AMD
$$ 

$$
\triangle OAB=\triangle OAM+ \triangle MAB= \triangle OAM+\triangle OAM+ \triangle OAD=$$ $$= \triangle OAM+\triangle OMC+\triangle OAD=\triangle OAC+\triangle OAD
$$
A: We have
$$\triangle{OAD}=\triangle{OCD}$$
and
$$\triangle{MCD}=\triangle{MAB}$$
So, 
$$\triangle{OAD}+\triangle{OAC}=\triangle{OCD}+\triangle{OAC}=\triangle{MCD}+\triangle{OAM}=\cdots$$
A: $$\triangle OAB=\triangle OAD+\triangle OAC$$
Means that
$$\triangle OAM + \triangle AMB = \triangle OAD+\triangle OAM + \triangle MOC$$
Means that
$$\triangle AMB = \triangle OAD+ \triangle MOC$$
But wait, $\triangle MOC = \triangle OAM$!
Observing that the line segment $\overline{AM}$ bisects $\triangle DAB$...
Can you finish?
A: Note: I will use the following notation $[\text{Figure}]$ for the area of the figure. Such as $[XYZ]$ denotes the area of $\triangle XYZ$.
Note that $\triangle ABD$ and $\triangle CDB$ are congruent, thus their areas are also equal, but since they share a common base $BD,$ the length of their altitudes $AQ$ and $CP$ must be equal for the areas of the triangles to be equal.

Thus we know that $\triangle AOD$ and $\triangle COD,$ which have the same base $OD$ are equal heights $AQ=CP,$ must have the same area. Thus, we have;
$$[OAD]=[OCD]$$
$$\implies [OAD]+[OAC]=[OCD]+[OAC]=[OACD].$$
Hence, the right hand side of the desired equation is the area of the quadrilateral $OACD.$

But quadrilateral $OACD$ can be bisected into $\triangle MCD$ and $\triangle OAM,$ so we have;
$$[OAD]+[OAC]=[OACD]=[MCD]+[OAM].$$
Since $\triangle MCD$ and $\triangle MAB$ are congruent, we know that $[MCD]=[MAB],$ and we have;
$$[OAD]+[OAC]=[MAB]+[OAM]=[OAB].$$
This is what was required.
A: Completing $AOC$ to a parallelogram centered at $M$, the diagram becomes a proof by dissection of the stated formula. 
The line segments in the completed diagram divide $OAB$ into $3$ pieces that can be assembled into the other two triangles.
For the formula to hold for all positions of $O$ the areas should be taken with $\pm$ signs according to the orientation of the triangles.
