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..
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
$$
\triangle AOB=\triangle AOM+\triangle AMD
$$ 



$$
\triangle OAB=\triangle OAM+ \triangle MAB= \triangle OAM+\triangle OAM+ \triangle OAD=$$ $$= \triangle OAM+\triangle OMC+\triangle OAD=\triangle OAC+\triangle OAD
$$
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.