On the rectangle $ABCD$, two points $M$ and $N$ have been marked on the sides $AB$ and $BC$ respectively.
The rectangle has lines drawn between $AN$, $ND$, $MC$, $MD$. It forms many different areas, many of which are triangles.
A triangle is formed in the lower left corner that has points $A$, $M$ and the point where the line $AN$ intersects $MD$ (call that point $E$). That triangle has area $3$.
Another triangle is formed in the upper left corner. It has points $C$, $N$ and the point where the line $ND$ intersects $CM$ (call that point $F$). That triangle has area $2$.
A quadrilateral is formed in the lower right corner. It has points $M$, $B$, $N$ and the point where the line $MC$ intersects $AN$ (call that point $G$). That quadrilateral has an area of $20$.
The goal of the assignment is to find the area of the quadrilateral $DEGF$.
You best understand the assignment when looking at it on paper, so I attached a image down below.
This question is from a high school entrance exam in Sweden.
I have tried to match triangles to each other with no success. I can guess that you would need a giant expression with a bunch of variables which eventually cancel out to get an answer, but I have gotten no success from that path. I also assume that you would need to figure out the length of AB and BC to solve the question. I have also looked at geometric identities but have gotten no success, and found none that seems relevant to the question.
Using the hint above, $(AMD)+(MBC) = R/2$, where $R$ is the area of the rectangle $ABCD, R=(ABCD)$. Likewise, $(ABN) + (NCD) = R/2$. Furthermore, $(AMD) = x +3, (MBC)=t+20+2,$ $(ABN) = 3+z+20, (NCD) = 2 + y,$ where $x$ is $(AED)$, $y$ is $(FDC)$, $z$ is $(AME)$ and $t$ is $(GNF)$.
Then, $(AMD)+(MBC)+ (ABN)+(NCD) = R =x+y+z+t + 52$, and $(DEFG) + x +y+z+t+3+20+2=R.$
Thus, $(DEFG) + R-52 + 27 = R, (DEFG) = 25.$