[proofplan] The proof computes the divisor of the rational function $L_0/L_1$ on $X$. Since the two linear forms have the same homogeneous degree and neither vanishes identically on $X$, this quotient defines a nonzero element of the function field $k(X)$. Along each prime divisor of the nonsingular variety $X$, its valuation is the vanishing order of $L_0$ minus the vanishing order of $L_1$, which is precisely the coefficient of $H_0|_X-H_1|_X$. Thus the difference of the two hyperplane sections is principal, so the two divisors are linearly equivalent. [/proofplan]

[step:Construct the quotient of the two hyperplane equations as a rational function on $X$]For $i\in\{0,1\}$, let $s_i$ denote the restriction of the homogeneous linear form $L_i$ to $X$, viewed as a rational section of the line bundle $\mathcal O_X(1)$. Because $H_i$ does not contain $X$, the section $s_i$ is not the zero rational section. Since $s_0$ and $s_1$ are rational sections of the same line bundle, their quotient defines a nonzero rational function \begin{align*} f:=\frac{s_0}{s_1}\in k(X)^\times. \end{align*} Equivalently, on every projective affine chart where a homogeneous coordinate is nonzero, this is the quotient of the two induced degree-zero regular functions, and these local quotients agree on overlaps because $L_0$ and $L_1$ have the same degree.[/step]

[guided]For each $i\in\{0,1\}$, the homogeneous linear form $L_i$ cuts out the hyperplane $H_i=V_+(L_i)$ in projective space. Restricting $L_i$ to $X$ gives a section $s_i$ of $\mathcal O_X(1)$. The hypothesis that $H_i$ does not contain $X$ means exactly that $L_i$ does not vanish identically on $X$, so $s_i$ is not the zero rational section. The quotient we want is \begin{align*} f:=\frac{s_0}{s_1}. \end{align*} This is a rational function on $X$ because $s_0$ and $s_1$ are sections of the same line bundle $\mathcal O_X(1)$. In local terms, after trivializing $\mathcal O_X(1)$ on a projective affine chart, both sections become regular functions, and their quotient is an ordinary rational function. On overlaps, the transition factors for $s_0$ and $s_1$ are the same, so they cancel in the quotient. Therefore these local quotients glue to a well-defined element \begin{align*} f=\frac{s_0}{s_1}\in k(X)^\times. \end{align*} The nonzero condition follows from the fact that neither $s_0$ nor $s_1$ is the zero rational section.[/guided]

[step:Compute the valuation of the quotient along each prime divisor] Let $E\subset X$ be a prime divisor, and let $\operatorname{ord}_E:k(X)^\times\to \mathbb Z$ denote the discrete valuation associated to the local ring of $X$ at the generic point of $E$. Since $X$ is nonsingular, this local ring is a discrete valuation ring. By additivity of $\operatorname{ord}_E$ on products and quotients in $k(X)^\times$, \begin{align*} \operatorname{ord}_E(f)=\operatorname{ord}_E(s_0)-\operatorname{ord}_E(s_1). \end{align*} By the definition of the hyperplane section divisor, the coefficient of $E$ in $H_i|_X$ is $\operatorname{ord}_E(s_i)$. Hence the coefficient of $E$ in $\operatorname{div}(f)$ is the coefficient of $E$ in $H_0|_X-H_1|_X$. [/step]

[step:Identify the difference of hyperplane sections as a principal divisor] The preceding valuation computation holds for every prime divisor $E\subset X$. Therefore the Weil divisors agree coefficientwise: \begin{align*} \operatorname{div}(f)=H_0|_X-H_1|_X. \end{align*} The right-hand side is therefore a principal divisor. By the definition of linear equivalence of divisors, two divisors are linearly equivalent precisely when their difference is principal. Thus \begin{align*} H_0|_X \sim H_1|_X. \end{align*} [/step]