Yes, following Eclid I.43 so I replace triangles with parallelogram and each area value is two times more.
Sorry for my bad explanation and later reply you (now morning in my place). For better understanding, I use notations in my original message.
One line through X and parallel with diagonal BD intersects two sides BC // DA at Cb, Ad respectively and intersects two sides AB // CD at Ab, Cd respectively. Then:
BDAdCb and BDCdAb are two diagonal parallelograms of X with respect to diagonal BD.
Of course:
Area(BDAdCb) = Area(BDCdAb) = 2*Area(BDX)
and we can formulate the result as:
Area(BDAdCb) = Area(BDCdAb) = Abs(Area(AMXQ) - Area(CNXP))

Best regards,
Bui Quang Tuan