How to prove that the sum of the areas of triangles $ABR$ and $ CDR$ triangle is equal to the $ADR$? In the convex quadrilateral $ABCD$, which is not a parallelogram, the line passing through the centers of the diagonals $AC$ and $BD$ intersects the segment $BC$ at $R$. How to prove that the sum of the areas of triangles $ABR$ and $CDR$ is equal to the area of triangle $ADR$? I have no idea how to do this. Can this be proved with simple geometry?


Let H and K be the midpoints of AC and BD respectively.
To prove the required, I will break the proof into several claims.
Claim #1: Every quadrilateral can be bisected (by a straight line) into two halves that are equal in area.
After joining AK and CK, ABCD is now divided into 4 triangles with [red] = [pink] and [blue] = [green].
Then, [pink] + [blue] = [white] + [green]. (See figure 1.) 

Construction: (1) Draw KE // AC cutting AB at E; and (2) Let CE cut AK at J. By “equal base, equal altitude” principle, [⊿AEJ] = [⊿CKJ]. This means we can transform ⊿AEJ to fit the position of ⊿CKJ such that [⊿BEC] = [quad AECD]. (See figure 3.)
 
After adding ⊿RAD, the next goal is to fill its area by pieces of quad AECD. ⊿RAD is partly filled by Quad APUD because it is common to both. Note that [⊿AEP] = [⊿TUC] + [quad QPUT]. The latter is used to cover part of ⊿RAD while ⊿TUC is to be connected to ⊿CUD to form the new ⊿TCD.
The remaining question is “will [⊿TCD] = [⊿RQT]?” We are done if we can prove DQ // BC.
Claim #2: BVDC is a parallelogram.

Further construction needed: (1) Extend CK to V such that AV // HK; (2) Join BV.
Applying intercept theorem to ⊿CAV, we have CK = KV. This and with the given BK = KD make BVDC a parallelogram.
