[step:Count translated blocks through a fixed pair of distinct points]Let $a,b \in G$ be distinct points. Define the set of indices of blocks containing both points by
\begin{align*}
I(a,b) := \{g \in G : a \in B_g \text{ and } b \in B_g\}.
\end{align*}
For $g \in G$, the conditions $a \in B_g$ and $b \in B_g$ are equivalent to the existence of elements $x,y \in D$ such that
\begin{align*}
a = g+x,
\qquad
b = g+y.
\end{align*}
For a fixed $g$, these elements are uniquely determined as $x=a-g$ and $y=b-g$. Therefore define the map $\Phi: I(a,b) \to \{(x,y) \in D \times D : b-a = y-x\}$ by sending each $g \in I(a,b)$ to $(a-g,b-g)$. This map is well-defined.
It is bijective. Indeed, if $g \in I(a,b)$, then $a-g,b-g \in D$ and
\begin{align*}
(b-g)-(a-g)=b-a,
\end{align*}
so $\Phi(g)$ lies in the displayed set. Conversely, if $(x,y) \in D \times D$ satisfies $b-a=y-x$, define $g:=a-x$. Then
\begin{align*}
g+y = a-x+y = a+(y-x)=a+(b-a)=b,
\end{align*}
and also $g+x=a$, so $a,b \in B_g$. Thus $g \in I(a,b)$ and $\Phi(g)=(x,y)$.
Since $a \ne b$, the element $b-a$ is nonzero. By the difference set hypothesis, there are exactly $\lambda$ ordered pairs $(x,y) \in D \times D$ such that $b-a=y-x$. Hence
\begin{align*}
|I(a,b)|=\lambda.
\end{align*}[/step]