Problem 799

56. Let the adjacent sides of the parallelogram be denoted by $a$ and $b$, and the vertices by $A, B, C, D$; thus, $A D=B C=a, A B=C D=b$. Let the point inside the parallelogram be denoted by $O$, the distance between $A D$ and $B C$ by $h_{1}$, and the distance between $A B$ and $C D$ by $h_{2}$. Then we have

$$S_{\triangle B O C}+S_{\triangle A O D}=\frac{1}{2} a h_{1} ; S_{\triangle A O B}+S_{\triangle C O D}=\frac{1}{2} b h_{2}, \text { but } a h_{1}=b h_{2} .$$

Therefore, $S_{\triangle B O C}+S_{\triangle A O D}=S_{\triangle A O B}+S_{\triangle C O D}$.