Parallelogram Proof

[BlacKKy]

"X" is a point within the triangle "DAC" of the parallelogram "ABCD". Prove that triangle "AXC" = triangle "AXB" - triangle "AXD"

In other word I have to prove that AXC=AXB-AXD)
Please help me to construct the proof, I think this question needs some constructions..
Below is the attachment of the image, please help...


EDIT: Ok I'll let you know some theorems also, so you guys can make the answer effective..
1) The parallelogram on the same base and between the same pair of parallel lines are equal in area.
2)If a triangle and a parallelogram are on the same base and between the same pair of parallel lines, then the area of the trainle is half the area of the parellogram
3)The triangles on the same base and between the same pair of parallel lines are equa in area.
4)The areas of triangles with a common vertex and with their bases lined on the same straight line are proportional to lengths of their bases.

 

[Freeman]

Let E be a point on AD such that EX//AB.
[AXC]+[AXD]=[ACD]-[XCD]=[ACD]-[ECD]=[ACE]=[ABE]=[AXB].

 

[BlacKKy]

Let E be a point on AD such that EX//AB.[AXC]+[AXD]=[ACD]-[XCD]=[ACD]-[ECD]=[ACE]=[ABE]=[AXB].

Thank you for your reply 1 question, how does it become ACD-XCD = ACD-ECD ??Even if I create a triangle ECD, I can't find a way to equalize ACD-XCD = ACD-ECD

 

[BlacKKy]

Thank you for your reply 1 question, how does it become ACD-XCD = ACD-ECD ??Even if I create a triangle ECD, I can't find a way to equalize ACD-XCD = ACD-ECD

Hey! Thank you! just now figured out the way!!! THANKS A BUNCH! I kept thinking about this question for about 6 hours but still didn't get the answer, if it wasn't you, I might never. THANK YOU A LOT!