Question on Proof that in a primitive group we have for the subdegrees $n_{i+1} \le n_i(n_2-1)$

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The following question is on a proof from Dixon & Mortimer, Permutation groups, page 72. He uses the characterisation that a transitive group $G$ acts primitive on $\Omega$ iff for each orbital its associated orbital graph is connected.

Lemma: Let $G$ be a finite primitive permutation subgroup of $Sym(\Omega)$ of degree $n$ and rank $r > 2$ with subdegrees $n_1 = 1 \le n_2 \le \ldots \le n_r$. Assume that $G$ is not regular. Then $n_{i+1} \le n_i(n_2-1)$ for all $i \ge 2$.

Proof: Fix $\alpha \in \Omega$ and order the orbitals $\Delta_1, \ldots, \Delta_r$ such that $|\Delta_i(\alpha)| = n_i$ for each $i$. To simplify notation, set $\Delta := \Delta_2$ and let $\Delta^*$ be its paired orbital. The set $\Delta \circ \Delta^*$ is $G$-invariant and consists of all pairs $(\alpha,\beta)$ such that $(\alpha, \gamma), (\beta, \gamma) \in \Delta$ for some $\gamma$. Moreover, $\Delta \circ \Delta^*$ contains a nondiagonal orbital because $n_2 > 1$. Now consider paths in the color graph of $G$ which start at $\alpha$ and have the form: $\alpha = \alpha_0, \alpha_1,\ldots, \alpha_k$ with each edge $(\alpha_i, \alpha_{i+1})$ in either $\Delta$ or $\Delta^*$ depending on whether $i$ is even or odd. We shall call such a pair an ''alternating path'' of length $k$. Since $\Delta \circ \Delta^*$ contains a nondiagonal orbits, the above mentioned characterisation of primitivity shows that for each $\beta \in \Omega$ there is an alternating path from $\alpha$ to $\beta$.

Suppose that $2 \le i < r$; we want to show that $n_{i+1} \le n_i(n_2 - 1)$. Let $k$ be the shortest length of a minimal alternating path from $\alpha$ to some vertex $\beta$ for which $(\alpha, \beta) \in \Delta_j$ and $j > i$. Fix such a path, $\alpha = \alpha_0, \alpha_1, \ldots, \ldots, \alpha_k = \beta$, and note that $k \ge 2$ because $j \ne i$. By the choice of $k$, $(\alpha, \alpha_{k-1}) \in \Delta_t$ for some $t \le i$. Now suppose that $k$ is odd [...].

I understood the proof, except the part I made bold above. Why do we must have $(\alpha, \alpha_{k-1}) \in \Delta_t$ for $t \le i$?