Dear All My Friends,
Two following triangle area facts in parallelogram and quadrilateral are very simple and equivalent:

1. Given any point X inside parallelogram ABCD.
One line from X parallel with AD//BC intersect AB at M, CD at P.
One line from X parallel with AB//CD intersect BC at N, DA at Q.
Result: Area(BDX) = Abs(Area(MQX) - Area(NPX))

2. X is intersection of two diagonals in convex quadrilateral MNPQ. E and F are midpoints of MN and PQ respectively.
Result: Area(EFX) = 1/4*Abs(Area(MQX) - Area (NPX))

Proof of 1)
Suppose X is inside triangle ABD.
Easy to show Area(MNPQ) = 1/2*Area(ABCD)
Area(BDX) = Area(ABD) - Area(MBX) - Area(QDX) - 2*Area(MQX) =
1/2*Area(ABCD) - Area(MNX) - Area(PQX) - 2*Area(MQX) =
Area(MNPQ) - Area(MNX) - Area(PQX) - 2*Area(QMX) =
Area(NPX) + Area(MQX) - 2*Area(MQX) =
Area(NPX) - Area(MQX)

Proof of 2)
Suppose X is inside MEFQ.
Area(EFX) = Area(EFY) - Area(EXY) - Area(FXY) =
1/4*Area(MNPQ) - 1/4*Area(MNQ) - 1/4*Area(MPQ) =
1/4*Area(MNPQ) - 1/4*(Area(MNQ) + Area(MPQ)) =
1/4* Area(MNPQ) - 1/4*(Area(MNPQ) - Area(NPX) + Area(MQX)) =
1/4*(Area(NPX) - Area(MQX))

Please kindly give me references if any one knows about these!

Thank you and best regards,
Bui Quang Tuan

This is indeed a nice generalization of Euclid I.43.

Dear Alex,

Thank you very much for your interesting reference! You are right.

Following Euclid I.43 we can rearrange this fact as following:

From any point inside a parallelogram we construct three lines: two lines parallel with the sides of the parallelogram and one line parallel with one diagonal.
Two lines parallel with the sides divide original parallelogram into four small parallelograms. If two small parallelograms have not any vertex on one diagonal we name they as complement parallelograms with respect to this diagonal.
One line parallel with one diagonal bound with this diagonal and two parallel sides of original parallelogram one new small parallelogram. We name this parallelogram as diagonal parallelogram. Of course there are two diagonal parallelograms with given diagonal but easy to show that they are always area equal.
Now we can formulate generalized Euclid I.43 as following:
In any parallelogram two complement parallelograms of given inside point with respect to given diagonal have area difference as area of respective diagonal parallelogram.
We get the Euclid I.43 if area of diagonal parallelogram is zero, i.e. given point is on the diagonal.
We can also formulate the fact for any point on the plane (inside and outside):
In any parallelogram three areas of two complement parallelograms and one respective diagonal parallelogram of given point with respect to given diagonal hold a property: one is sum of other two.

Thanks to Euclid we have our geometry now!
Best regards,
Bui Quang Tuan

Could you please expand on this please. I do not see what a diagonal parallelogram is. Thank you.

Ah, I got it. Just replace the triangles with paralellograms and use some shear. Like it.

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

Another view:
http://www.cut-the-knot.org/Curriculum/Geometry/EuclidI43Extended.shtml

(preliminary)

Red and blue areas are equal.

Dear Alex,
Yes, it is another interesting variant.
In your drawing, for symmetry we can see some more equal area parallelograms:
U = intersection of BY, MP
V = intersection of AD with line from U and parallel with AB // CD
1. Area(CNXP) = Area(AQRS) = Area(AMUV)
2. Area(XBYD) = Area(MSRX) = Area(QVUX)
Best regards,
Bui Quang Tuan

Yes, that's right.

What I like about this variant is that the areas do not intersect. So that the illustration does not require a lot of explanations.

Dear Alex,
I recommend two proofs for your variant.
W = intersection of DR and AB

1. Proof used my variant 1)
As my variant:
(XNCP) - (AMXQ) = 2*(BXD) = (XBYD) = (XRWB) = (XRSM)
From this:
(XNCP) = (AMXQ) + (XRSM) = (ASRQ)

2. Proof used Euclid I.43
A line from W parallel with BC intersects NQ at E, CD at F.
K is intersection of RS and CD
RE = BW - NR = XR - NR = XN there for two paralellograms CNXP and FERK are congruent.
By Euclid I.43: (ASRQ) = (FERK) = (XNCP)

Also please check to see some typo mistakes in your new page:
http://www.cut-the-knot.org/Curriculum/Geometry/EuclidI43Extended.shtml

In the first line:
"Through point X inside the parallelogram ABCD draw lines MP||CD and NQ||AB."
I think MP is parallel with BC.

At the end of the page:
"This means (and follows from the fact) that the point of intersection of RS and UV (not shown) lies on the line AX, the diagonal of parallelogram ASXR."
I think there is one typo mistake because ASXR is not a parallelogram.

Best regards,
Bui Quang Tuan

Thank you.