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?

 A: 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 $$[DAR] = [ABCD]-([ABR]+[CDR])=[ABCD] - \frac 12 [ABCD]=\frac 12[ABCD] = [ABR]+[CDR].$$
A: 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.
A: 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.
A: I am sorry, I withdraw (not completely ; see at the bottom) what I have said. I had read "line $AB$" instead of "segment $AB$". 
Nevertheless, I think that, instead of "the line passing..." one should have found "in the case where the line passing through the midpoints $I$ et $J$ intersects segment $[BC]$, then ...".
I have looked for a geometric proof, but found none. Here is an analytical proof.
Take coordinate axis such that $B(-1,0)$ and $C(1,0)$ (without loss of generality). Let $A(a,b)$ and $D(c,d)$. Let $R(r,0)$.
The collinearity condition between $R$ and $I,J$ is
$$\begin{vmatrix}r&\frac{a+1}{2}&\frac{c-1}{2}\\
0&\frac b2&\frac d2\\ 1&1&1\end{vmatrix}=0 \ \ \Rightarrow \ \ r=\dfrac{b+d+ad-bc}{2(d-b)} \ \ (1)$$ 
Now, it is a matter of (rather simple) computation to check that the difference of areas:
$$\delta=A(ABR)+A(CDR)-A(ADR)$$ is zero. Indeed,
$$\delta=\frac12det(\vec{RB},\vec{RA})+\frac12det(\vec{RC},\vec{RD})-\frac12det(\vec{RD},\vec{RA})$$
which is equivalent (after multiplication by $2$) to:
$$2\delta=\begin{vmatrix}a-r&-1-r\\b&0\end{vmatrix}+\begin{vmatrix}1-r&c-r\\0&d\end{vmatrix}+\begin{vmatrix}a-r&c-r\\b&d\end{vmatrix}$$
Expanding previous expression and replacing by the value of $r$ given by (1)   gives $\delta=0$, as desired.
Note that a condition of existence of $R$ is $b \neq d$, i.e., trapezoids are not accepted (one can check indeed that, in such a case, line $IJ$ is parallel to $BC$).
Remark: The result, true for $R \in [BC]$, is no longer true if the line joining the midpoints of the diagonal intersect line BC outside segment [BC], if we continue to work with unsigned areas.
Here is a counterexample.
Let $A(0,2), B(1,0),C(2,0),D(3,1)$. 
This gives for the midpoints: $I(1,1)$ and $J(2,1/2)$. Thus $R(3,0)$.
The areas are $A(ABR)=2, A(CDR)=1/2$ whereas $A(ADR)=3/2$ instead of $5/2$.
