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 is tricky. 

We shall use an easy fact that if we are given fixed points $Y,Z$, and a variable point $X$ that changes linearly, then $[XYZ]$ changes linearly, where $[\mathcal{F}]$ denotes the oriented area of $\mathcal{F}$. 

Let $M,N$ be midpoints of $AC, BD$. Using the fact we know that the function $MN \ni X \mapsto [ABX]+[CDX]$ is linear. However $$[ABN]+[CDN]=\frac 12 [ABD] + \frac 12 [CDB] = \frac 12 [ABCD]$$ and $$[ABM]+[CDM]=\frac12 [ABC] + \frac 12 [CDA] = \frac 12 [ABCD]$$ so this function is actually constant. In particular $$[ABR]+[CDR]=\frac 12 [ABCD].$$ This implies that $$[DAR] = [ABCD]-([ABR]+[CDR])=[ABCD] - \frac 12 [ABCD]=\frac 12[ABCD] = [ABR]+[CDR].$$
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 &amp; CDR &amp; ADR and solved for a common relationship.  I recommend finding another way.