Let $A,B,C,D$ be the vertices of a convex quadrilateral; $K,L,M,N$ are the mid points of the sides; $X,Y,Z,T$ are intersections of the line joining vertices to mid points of opposite sides, and $a,b,c,d$ are areas of portions of the quadrilateral. Find the area of $XYZT$ in terms of $a,b,c,d$.
I could join $BD$ and $AC$ and also $LD$ and $KC$, and thereafter applied 'median divides a triangle into two equal halve in terms of area.' But it led me nowhere.
$$S_{\Delta ABL}=\frac{1}{2}S_{\Delta ABC}$$ and
$$S_{\Delta NCD}=\frac{1}{2}S_{\Delta ACD}.$$ Thus, $$S_{\Delta ABL}+S_{\Delta NCD}=\frac{1}{2}S_{\Delta ABC}+\frac{1}{2}S_{\Delta ACD}=\frac{1}{2}S_{ABCD}.$$ Hence, $$S_{ALCN}=\frac{1}{2}S_{ABCD}.$$ By the same way we'll obtain: $$S_{KBMD}=\frac{1}{2}S_{ABCD}.$$ Thus, $$0=S_{ALCN}+S_{KBMD}-S_{ABCD}=S_{XYZT}-a-b-c-d,$$ which gives $$S_{XYZT}=a+b+c+d.$$