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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.

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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 $\,\Delta BCX\,\,,\,\Delta CXD\,$

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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 $AXCB)$

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