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?


This link gives a great visual representation of this proof.
https://www.mathsisfun.com/geometry/quadrilaterals-interactive.html.  
Analytically, I created a complicated proof by using the equation base*height/2 = area_of_a_triangle and cos(angle)*height = base for each of ADR & CDR & ADR and solved for a common relationship.  I recommend finding another way.
