Let $M$ be a point lie inside $ABCD$ such that areas of $ABCM$ and $AMCD$ are equal.
Case $1$: $M$ lie on the half plane $AC$ contain $D$ Let $X$ be the reflection of $B$ over $M$. Since $MB=MX$ so $A_{AMCX}=A_{AMX}+A_{CMX}=A_{ABM}+A_{CBM}=A_{ABCM}=\frac{A_{ABCD}}{2}=A_{AMCD}$
So that:
$A_{ADC}=A_{AMCD}+A_{MAC}=A_{AMCX}+A_{MAC}=A_{ACX}$
Let $F,G$ be the projections of $D,X$ on $AC$.
We have: $A_{ADC}=\frac{DF*AC}{2}=A_{ACX}=\frac{XG*AC}{2}$
So that $DF=XG$ so $DX//AC$ $=>$ $M$ lie on the line from mid point $E$ of $BD$ parallel to $AC$. For case $2$ also the same.
