5.1. In parallelogram $O A B C$, the vertices $O(0 ; 0 ;)$, $A(3 ; 6)$, and $B(8 ; 6)$ are given. Find the ratio of the lengths of the diagonals $O B$ and $A C$, and also write the equations of the sides of the parallelogram and the diagonal $A C$.
5.1. Since the ordinates of vertices $A$ and $B$ are equal, $A B \| O x$ (Fig. 5.1). Of the three segments $O A, A B$, and $O B$, only $O A$ and $A B$ can be the sides of the parallelogram, as $O B$ is the diagonal by condition; therefore, $B C \| O A$ and $C(5 ; 0)$. Using formula (5.1), we find

$O B=\sqrt{8^{2}+6^{2}}=\sqrt{100}, A C=\sqrt{(5-3)^{2}+(0-6)^{2}}=\sqrt{40}$; thus, $O B: A C=\sqrt{100}: \sqrt{40}=\sqrt{2.5}$ - the required ratio of the diagonals.

According to formula (5.3), the equation of side $O A$ is $y=k x+q$, where $k=6: 3=2$ and $q=0$; therefore, $y=2 x$. Using equality (5.7), we write the equation of side $A B: y=6$. Further, since $B C \| O A$, the slope of line $B C$ is $k=2$ by formula (5.8), and the corresponding value of $q$ is found from the equation $y=2 x+q$ by substituting the coordinates of point $C(5 ; 0)$ for $x$ and $y$; then we get $0=10+q$, i.e., $q=-10$; hence, the equation of $B C$ is $y=2 x-10$. Finally, the equation of $O C$ is $y=0$.

To find the equation of diagonal $A C$, we use the fact that points $A(3 ; 6)$ and $C(5 ; 0)$ belong to the line $A C$ and, therefore, their coordinates satisfy the desired equation. Substituting these coordinates into the equation $y=k x+q$, we get $6=3 k+q, 0=5 k+q$, from which $k=-3, q=15$. Thus, $y=-3 x+15$ is the equation of diagonal $A C$.

Answer: $\quad O B: A C=\sqrt{2.5} ; y=0(O C), y=6(A B), y=2 x(O A)$, $y=2 x-10(B C) ; y=-3 x+15(A C)$.