[proofplan] Fix a point $x \in X$ and count, in two ways, the finite set of incident pairs consisting of another point $y$ and a block $B$ containing both $x$ and $y$. Counting first by the point $y$ gives $\lambda(v-1)$ such pairs, because each point distinct from $x$ lies with $x$ in exactly $\lambda$ blocks. Counting first by the block $B$ gives $r_x(k-1)$ such pairs, because each block through $x$ contains exactly $k-1$ further points. Equating the two counts gives the formula for $r_x$, and the right-hand side does not depend on $x$. [/proofplan] [step:Count pairs by first choosing the second point] Fix a point $x \in X$. Define the finite set \begin{align*} S_x := \{(y,B) \in X \times \mathcal{B} : y \neq x,\ (x,B) \in I,\ (y,B) \in I\}. \end{align*} For each point $y \in X \setminus \{x\}$, the defining property of a $2-(v,k,\lambda)$ design says that the two distinct points $x$ and $y$ are jointly incident with exactly $\lambda$ blocks. Since $|X \setminus \{x\}| = v-1$, summing over the possible choices of $y$ gives \begin{align*} |S_x| = \lambda(v-1). \end{align*} [guided] We fix $x \in X$ because the goal is to compute the number of blocks through this particular point and then show that the answer does not depend on which point was chosen. The object we count is \begin{align*} S_x := \{(y,B) \in X \times \mathcal{B} : y \neq x,\ (x,B) \in I,\ (y,B) \in I\}. \end{align*} Thus an element of $S_x$ consists of a point $y$ different from $x$ and a block $B$ incident with both $x$ and $y$. Now choose $y$ first. There are exactly $v-1$ possible choices for $y$, because $|X|=v$ and the condition $y \neq x$ removes one point. For each such $y$, the points $x$ and $y$ are distinct, so the defining property of a $2-(v,k,\lambda)$ design applies: exactly $\lambda$ blocks are incident with both of them. Therefore the number of pairs in $S_x$ is \begin{align*} |S_x| = \sum_{y \in X \setminus \{x\}} \lambda = \lambda |X \setminus \{x\}| = \lambda(v-1). \end{align*} [/guided] [/step] [step:Count the same pairs by first choosing a block through $x$] Let \begin{align*} \mathcal{B}_x := \{B \in \mathcal{B} : (x,B) \in I\} \end{align*} be the set of blocks incident with $x$, so that $|\mathcal{B}_x|=r_x$. If $B \in \mathcal{B}_x$, then $B$ is incident with exactly $k$ points, one of which is $x$. Hence exactly $k-1$ points $y \in X \setminus \{x\}$ satisfy $(y,B) \in I$. Counting $S_x$ by first choosing $B \in \mathcal{B}_x$ gives \begin{align*} |S_x| = r_x(k-1). \end{align*} [/step] [step:Equate the two counts and identify the common replication number] The two preceding counts are counts of the same finite set $S_x$, so \begin{align*} r_x(k-1) = \lambda(v-1). \end{align*} Since $k \geq 2$, we have $k-1 \neq 0$, and division by $k-1$ gives \begin{align*} r_x = \frac{\lambda(v-1)}{k-1}. \end{align*} The right-hand side depends only on $v$, $k$, and $\lambda$, not on the chosen point $x$. Therefore every point has the same replication number, with common value \begin{align*} r = \frac{\lambda(v-1)}{k-1}. \end{align*} [/step]