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.