If I am given a quadrilateral, how can I prove that by there exists a point $P$ in the interior of the quadrilateral so that one can draw straight lines from this point $P$ to all the vertices of the quadrilateral (these do not necessarily form diagonals) in a manner that this produces four triangles all of whom have equal areas if and only if two edges of the quadrilateral are parallel lines.
First, I am starting by assuming that all areas of the triangles are equal to $1$. Now, let $b_k$ be the base of triangle $k$ for $k\in\{1,2,3,4\}$ and $h_k$ be the height. Then clearly $\frac{1}{2}h_kb_k=1$ for each $k\in \{1,2,3,4\}$. However, we do not have sufficient information only by this. I think a geometric argument would be needed.
EDIT: seems like my initial claim is false. It would be interesting to see if this is in fact possible for all convex quadrilaterals and if it is not, then for what type of convex quadrilaterals does the claim fail for?
Best Answer
Inside a convex quadrilateral $ABCD$ there exists a point $P$ such that triangles $ABP$, $BCP$, $CDP$, $DAP$ have the same area, if and only if $ABCD$ is split by one of its diagonals into two triangles with the same area.
The "if" part of the proof is obvious: just take as $P$ the midpoint of that diagonal.
To prove the "only if" part, suppose $P$ exists as stated above. If $P$ lies on diagonal $AC$, then $ABC$ and $ADC$ have the same area and we are done. Otherwise, let $M$ be the intersection between line $BP$ and diagonal $AC$. Points $A$ and $C$ have the same distance from line $BP$, hence triangles $BMA$ and $BMC$ have the same area and $M$ is the midpoint of $AC$. The same argument can be repeated with line $DP$: it must intersect $AC$ at its midpoint $M$. Hence $DPB$ are aligned and triangles $ABD$, $BCD$ have the same area. That completes the proof.
EDIT.
If quadrilateral $ABCD$ is not convex, the above statement still holds and the proof still works. In fact, suppose $D$ is the vertex lying inside the triangle formed by the other vertices (figure below). In this case diagonal $AC$ lies outside the triangle, so the only diagonal which can divide the quadrilateral into two triangles with the same area is $BD$. In addition, point $P$ must lie inside the convex quadrilateral formed by sides $BA$ and $BC$ with the extensions of $AD$ and $CD$, otherwise the segments connecting $P$ with the vertices would partially lie outside quadrilateral $ABCD$. This implies that lines $BP$, $DP$ both intersect diagonal $AC$ and the proof outlined above can be repeated as it is.