Lemma 1.19 in Conway's Complex Analysis Book

I am working through the following lemma in Conway's book on Complex Analysis (see page 65):

1.19 Lemma. If $G$ is an open set in $\Bbb{C}$, $\gamma : [a,b] \to G$ is a rectifiable path, and $f : G \to \Bbb{C}$ is continuous, then for every $\epsilon > 0$, there is a polygonal path $\Gamma$ in $G$ such that $\Gamma (a) = \gamma (a)$, $\Gamma (b) = \gamma (b)$, and $|\int_{\gamma} f - \int_{\Gamma} f| < \epsilon$.

In the proof, they construct $\Gamma$ by choosing a partition $\{t_0 < t_1 < ... < t_n\}$ of $[a,b]$ such that $$|\gamma (s) - \gamma (t)| < \delta ~~~~~~~~~~(*)$$ if $t_{k-1} \le s,t \le t_k$; and such that for $t_{k-1} \le \tau_k \le t_k$ we have $$\left| \int_{\gamma} f - \sum_{k=1}^{n} f(\gamma (\tau_k)) [\gamma (t_k) - \gamma (t_{k-1})] \right| < \epsilon.$$ With this, Conway defines $\Gamma : [a,b] \to \Bbb{C}$ by $$\Gamma (t) = \frac{1}{t_k - t_{k-1}} [(t_k - t) \gamma (t_{k-1}) + (t-t_{k-1})\gamma (t_k)]$$ if $t_{k-1} \le t \le t_k$. He then claims that it follows from (*) that $$|\Gamma (t) - \gamma (\tau_{k})| < \delta$$ for $t_{k-1} \le t \le t_{k}$. However, I am having trouble seeing why this inequality holds. I could use some help.

Note that, by multiplying by $1 = \frac{t_k - t + t - t_{k - 1}}{t_k - t_{k - 1}}$, $$ \gamma(\tau_k) = \frac{t_k - t}{t_k - t_{k - 1}} \gamma(\tau_k) + \frac{t - t_{k - 1}}{t_k - t_{k - 1}} \gamma(\tau_k). $$ Thus (some triangle inequality business and $(*)$), $$ \begin{align*} | \Gamma(t) - \gamma(\tau_k) | &= \Bigl | \frac{t_k - t}{t_k - t_{k - 1}} (\gamma(t_{k - 1}) - \gamma(\tau_k)) + \frac{t - t_{k - 1}}{t_k - t_{k - 1}} (\gamma(t_{k}) - \gamma(\tau_k)) \Bigr | \\ &< \frac{t_k - t + t - t_{k - 1}}{t_k - t_{k - 1}} \delta = \delta. \end{align*} $$