[proofplan] We prove the standard incidence-counting identity for this design directly, avoiding any external citation. The idea is to count the same set of incident point-block pairs in two ways: first by summing over blocks, and then by summing over points. Symmetry gives $|\mathcal B| = |X| = v$, so the resulting equality reduces to $vk = vr$, and cancellation by the positive integer $v$ gives $k = r$. [/proofplan] [step:Count point-block incidences by summing over the blocks] Define the incidence set $I$ by \begin{align*} I := \{(x,B) \in X \times \mathcal B : x \in B\}. \end{align*} For each block $B \in \mathcal B$, the hypothesis of a $2$-$(v,k,\lambda)$ design gives $|B| = k$. Since $I$ is the disjoint union of the sets $\{(x,B) : x \in B\}$ as $B$ ranges over $\mathcal B$, we obtain \begin{align*} |I| = \sum_{B \in \mathcal B} |B| = \sum_{B \in \mathcal B} k = |\mathcal B|\,k. \end{align*} [/step] [step:Count point-block incidences by summing over the points] For each point $x \in X$, the replication number $r$ is defined so that $x$ lies in exactly $r$ blocks. Equivalently, \begin{align*} |\{B \in \mathcal B : x \in B\}| = r. \end{align*} Since $I$ is also the disjoint union of the sets $\{(x,B) : B \in \mathcal B,\ x \in B\}$ as $x$ ranges over $X$, we obtain \begin{align*} |I| = \sum_{x \in X} |\{B \in \mathcal B : x \in B\}| = \sum_{x \in X} r = |X|\,r = vr. \end{align*} [guided] We are counting the same incidence set \begin{align*} I := \{(x,B) \in X \times \mathcal B : x \in B\} \end{align*} from the point side rather than from the block side. Fix a point $x \in X$. The replication number $r$ means exactly that the collection of blocks containing $x$ has cardinality $r$: \begin{align*} |\{B \in \mathcal B : x \in B\}| = r. \end{align*} Now partition $I$ according to its first coordinate. For distinct points $x,y \in X$, the sets of incidences with first coordinate $x$ and first coordinate $y$ are disjoint, and every incidence pair has exactly one first coordinate. Therefore \begin{align*} |I| = \sum_{x \in X} |\{B \in \mathcal B : x \in B\}| = \sum_{x \in X} r = |X|\,r. \end{align*} Since the design has $v$ points, $|X| = v$, so this becomes \begin{align*} |I| = vr. \end{align*} This is the same incidence set counted in the previous step, where summing over blocks gave $|I| = |\mathcal B|k$. [/guided] [/step] [step:Use symmetry to cancel the number of points] The two incidence counts give \begin{align*} |\mathcal B|\,k = vr. \end{align*} Because $\mathcal D$ is symmetric, $|\mathcal B| = v$, so \begin{align*} vk = vr. \end{align*} Since $X$ has $v$ points and a $2$-design has a nonempty point set, $v > 0$. Cancelling the positive integer $v$ gives \begin{align*} k = r. \end{align*} Thus $r = k$, as required. [/step]