[proofplan]
We introduce the two integer parameters forced by the design: the number $r_x$ of blocks through a point $x$, and the total number $b$ of blocks. First we count incident pairs $(y,B)$ with $y \neq x$ and $\{x,y\} \subset B$ to prove that $r_x$ is independent of $x$ and satisfies $r(k-1)=\lambda(v-1)$. Then we count point-block incidences in two ways to obtain $vr=bk$. Combining these two integer equations gives the two divisibility conditions.
[/proofplan]
[step:Count blocks through one fixed point]
Let $(X,\mathcal{B})$ be a $2-(v,k,\lambda)$ design. For each point $x \in X$, define
\begin{align*}
r_x := |\{B \in \mathcal{B} : x \in B\}|,
\end{align*}
the number of blocks containing $x$.
Fix $x \in X$. Define the set of pointed incidences away from $x$ by
\begin{align*}
I_x := \{(y,B) \in (X \setminus \{x\}) \times \mathcal{B} : x \in B \text{ and } y \in B\}.
\end{align*}
We count $I_x$ in two ways. For each $y \in X \setminus \{x\}$, the pair $\{x,y\}$ is contained in exactly $\lambda$ blocks, so counting first by $y$ gives
\begin{align*}
|I_x| = \lambda(v-1).
\end{align*}
Counting first by blocks containing $x$, each such block $B$ contributes exactly $k-1$ choices of $y \in B \setminus \{x\}$, so
\begin{align*}
|I_x| = r_x(k-1).
\end{align*}
Therefore
\begin{align*}
r_x(k-1)=\lambda(v-1).
\end{align*}
The right-hand side does not depend on $x$, so $r_x$ is the same integer for all $x \in X$. Denote this common value by $r \in \mathbb{N}$. Since $r(k-1)=\lambda(v-1)$, we have
\begin{align*}
k-1 \mid \lambda(v-1).
\end{align*}
[guided]
Fix a point $x \in X$. The goal is to extract an integer divisibility condition from the design axiom. Define
\begin{align*}
r_x := |\{B \in \mathcal{B} : x \in B\}|,
\end{align*}
so $r_x$ is the number of blocks passing through the fixed point $x$.
To relate $r_x$ to $\lambda$, we count the same finite set in two different ways. Define
\begin{align*}
I_x := \{(y,B) \in (X \setminus \{x\}) \times \mathcal{B} : x \in B \text{ and } y \in B\}.
\end{align*}
An element of $I_x$ records a second point $y \neq x$ together with a block $B$ containing both $x$ and $y$.
First count by the choice of $y$. There are $v-1$ possible points $y \in X \setminus \{x\}$. For each such $y$, the two-element subset $\{x,y\}$ is contained in exactly $\lambda$ blocks by the definition of a $2-(v,k,\lambda)$ design. Hence
\begin{align*}
|I_x|=\lambda(v-1).
\end{align*}
Now count by the choice of the block $B$. There are $r_x$ blocks containing $x$. If $B$ is one of them, then $|B|=k$, and one point of $B$ is already fixed as $x$. Therefore there are exactly $k-1$ choices of $y \in B \setminus \{x\}$. Hence
\begin{align*}
|I_x|=r_x(k-1).
\end{align*}
Equating the two counts gives
\begin{align*}
r_x(k-1)=\lambda(v-1).
\end{align*}
This equation also shows that $r_x$ is independent of $x$, because the right-hand side depends only on $v,k,\lambda$. Denote the common value by $r \in \mathbb{N}$. Since
\begin{align*}
r(k-1)=\lambda(v-1)
\end{align*}
with $r$ an integer, the integer $k-1$ divides $\lambda(v-1)$:
\begin{align*}
k-1 \mid \lambda(v-1).
\end{align*}
[/guided]
[/step]
[step:Count all point-block incidences]
Define
\begin{align*}
b := |\mathcal{B}|,
\end{align*}
the number of blocks of the design. Define the incidence set
\begin{align*}
I := \{(x,B) \in X \times \mathcal{B} : x \in B\}.
\end{align*}
Counting first by points, each $x \in X$ lies in exactly $r$ blocks, so
\begin{align*}
|I| = vr.
\end{align*}
Counting first by blocks, each $B \in \mathcal{B}$ contains exactly $k$ points, so
\begin{align*}
|I| = bk.
\end{align*}
Thus
\begin{align*}
vr=bk.
\end{align*}
[/step]
[step:Combine the incidence equations to obtain the second divisibility condition]
From the first count,
\begin{align*}
r=\frac{\lambda(v-1)}{k-1}.
\end{align*}
Substituting this value of $r$ into $vr=bk$ gives
\begin{align*}
b
= \frac{vr}{k}
= \frac{v}{k}\cdot \frac{\lambda(v-1)}{k-1}
= \frac{\lambda v(v-1)}{k(k-1)}.
\end{align*}
Since $b=|\mathcal{B}|$ is an integer, the numerator is divisible by the denominator:
\begin{align*}
k(k-1) \mid \lambda v(v-1).
\end{align*}
Together with the first divisibility condition, this proves the theorem.
[/step]
