## Task 27/66

What conditions must the diagonals of a plane convex quadrilateral $ABCD$ with the diagonal intersection point $S$ satisfy if the sum of the areas of triangles $ABS$ and $DSC$ is to be equal to the sum of the areas of triangles $BCS$ and $ASD$?
If the diagonal segments are denoted as follows: $A S=w ; B S=x ; C S=y ; D S=z$, and further $F(A B S)=A_{1} ; F(B S C)=A_{2} ; F(C S D)=A_{3} ; F(D S A)=A_{4}$ and $\measuredangle A S B=\rho$, then we have

$$
\begin{array}{rlrl}
A_{1} & =\frac{1}{2} w x \sin \rho & ; & A_{2}=\frac{1}{2} x y \sin 180^{\circ}-\rho=\frac{1}{2} x y \sin \rho \\
A_{3} & =\frac{1}{2} y z \sin \rho \quad ; & A_{4}=\frac{1}{2} w z \sin 180^{\circ}-\rho=\frac{1}{2} w z \sin \rho \\
A_{1}+A_{3} & =\frac{1}{2} \sin \rho(w x+y z) & ; \quad A_{2}+A_{4}=\frac{1}{2} \sin \rho(x y+w z)
\end{array}
$$

If $A_{1}+A_{3}=A_{2}+A_{4}$, then it follows that $w x+y z=x y+w z$ must hold (since $\rho \neq 0^{\circ}$ and $\rho \neq 180^{\circ}$, it also follows that $\sin \rho \neq 0$). A rearrangement yields $(w-y)(x-z)=0$.

This equation is satisfied exactly when $w=y$ or when $x=z$, i.e., when at least one of the diagonals is bisected by $S$.

From $A_{1}+A_{3}=A_{2}+A_{4}$, it follows that $w=y$ or $x=z$. Since all operations performed are uniquely reversible, it follows conversely that if $w=y$ or $x=z$, then $A_{1}+A_{3}=A_{2}+A_{4}$. The condition that at least one of the diagonals bisects the other is therefore both necessary and sufficient for the two area sums to be equal.