Relation between areas of two quadrilaterals If it is given that:
in a  quadrilateral $ABCD$, $X$ is the mid point of the diagonal $BD$, then prove that
area of $AXCB$ = $\frac12$ area of $ABCD$.
I do not know where to start, but I think we can take two cases, something like: 
$1$. When $AXCB$ is quadrilateral.
$2$. When $AXCB$ is a triangle, like in case when ABCD is parallelogram. 
Now, I want to ask, is this approach correct? Maybe it is long, but is it right? 
If it is, then could anyone suggest a shorter and more elegant solution? If it is not then what can I think of, here? like any theorem or any formula or any thing.
Well, rather an elegant solution I would prefer to understand the problem which I think I am not able to do. 
 A: Draw a diagram, with $\,A\,$ the upper left vertex of $\,ABCD\,$ and go clockwise, mark point $\,X\,$ on $\,BD\,$ and draw $\,AX\,\,,\,XC\,$.
After a moment of thinking, get convinced that in triangle $\,\Delta AXB\,$ , the height to $\,BX\,$ is exactly the same as the height to $\,DX\,$ in triangle $\,\Delta AXD\,$ , and thus $\,S_{\Delta ADX}=S_{\Delta ABX}\,$ , and exactly the same argument works for the other pair of triangles $\,\Delta BCX\,\,,\,\Delta CXD\,$
A: Since $X$ is mid point of $BD, BX = XD$
Get $Y, Z$ points such  that $BD⊥AY$ and $BD⊥CZ$.
$ AREA$ of $AXCB$ = $AREA$ of $∆AXB$ + $AREA$ of $∆CXB$
              = $\frac{1}{2}  BX . AY  + \frac{1}{2}  BX . CZ$
              = $\frac{1}{2}  BX .(AY + CZ)$
$AREA$ of $ABCD$ = $AREA$ of$ ∆ABD + AREA$ of $∆CBD$=
                 $\frac{1}{2}  BD . AY  + \frac{1}{2}  BD . CZ$
                  = $\frac{1}{2}  BD . (AY + CZ)$
                  =$\frac{1}{2} . 2BX . (AY + CZ) $
                  =$2 . (\frac{1}{2}  BX . (AY + CZ))$
                  = $2 .( AREA$ of $AXCB)$
