Uniform convergence of constant speed $C^1$ curves with the same endpoints This is an exercise which has been asked here also: 

Let $a,b\in\mathbb{R}^{2}$. Let $\{\sigma_n\}_{n=1}^\infty$ be a sequence curves $
\sigma_n:[0,1]\to{\Bbb R^2}
$
  such that $$\sigma_n(0)=a,\ \ \sigma_n(1)=b,$$ and $$\sigma_n\in C^1([0,1])$$ for all $n$. Also, assume that 
  $$
\forall t\in[0,1],\quad \|\sigma_n'(t)\|=L_n
$$
  and $$
\lim_n L_n=\|b-a\|
$$
  where $\|\cdot\|$ is the Euclidean norm for ${\Bbb R^2}$. Show that $\{\sigma_n\}$  uniformly converges to 
  $$
\sigma(t)=a+t(b-a). 
$$

I have one more question to this exercise (while I don't have a proof for the statement though):

Is $\{\sigma_n'\}$ also uniformly convergent?

I think a proof to the exercise I cited might help. (For the sake of a proof of the exercise: the boundedness of $\{L_n\}$ imply by the Arzelà–Ascoli theorem that $\{\sigma_n\}$ has a uniformly convergent subsequence. But I don't see how can I go on. )
 A: Hint: 
Use the fact that in a compact subset of $\Bbb R^n$ (here $I=[0,1]$) if the family {$σ_n$} is equicontinuous and converges pointwise to a function (here $σ$) ,which is necessarily continuous, then the convergence is uniform on $I$.
Show:
1)The family {$σ_n$} is equicontinuous because all it's elements have derivatives bounded by the same Lipschitz constant.
2)$\lim_n L_n=\|b-a\|=\|σ'(t)\|<=>$for every $ε>0$ there is a $n_0$ :$\|L_n-\|σ'(t)\|\|<ε ,n\geq n_0$
3)Draw a circle with it's centre on the line of $σ(t)$ (so  $σ(t)$ is it's diametre) and with radius $\|b-a\|/2$.
Let $G_{n-n_0}$={$σ_n(t):t\in I,\|b-a\|\leq \|σ'_n(t)\|\leq \frac {π\|b-a\|}{n-n_0}$}.
Then the family {$G_i$} is a decreasing sequence of closed sets and {$σ(t)$}$\subset \cap G_i$. $σ(t)$ is the only curve in $\Bbb R^2$ with lenght $\|b-a\|$ and end points $a,b$ ,it's is also an
accumulation point of {$G_i$} thus $σ_n(t)\to σ(t)$.
4)We have that $σ_n(t)\to σ(t)$ and that the family {$σ_n$} is equicontinuous. Use Hint
Proof(Hint):
Let ε>0. Because {$σ_n$} is equicontinuous ,for every $y\in I$ there is a $δ_{y}$ :for every $k$ 
i)$x\in I$ and $\|x-y\|<δ_{y}=>\|σ_{k}(x)-σ_{k}(y)\|<ε$
by letting $k\to \infty$ we have 
ii)$x\in I$ and $\|x-y\|<δ_{y}=>\|σ(x)-σ(y)\|\leqε$.
The family {$B(y,δ_{y}):y\in I$} is an open cover of $I$ and thus there exist $y_1,y_2,...,y_n\in I: I\subset B(y_1,δ_{y_1})\cup B(y_2,δ_{y_2})\cup...\cup B(y_n,δ_{y_n})$.
Let $k_0\in \Bbb N$: for every $j=1,2,..,n$
iii)$k\geq k_0=>\|σ_{k}(y_j)-σ(y_j)\|<ε$.
Let now $x\in I$. Then there is a $j=j(x)$ :$x\in B(y_j,δ_{y_j})$ 
and thus from i),ii),iii) we have that for every $k\geq k_0$, 
$\|σ_{k}(x)-σ(x)\|\leq \|σ_{k}(x)-σ_{k}(y_j)\|+\|σ_{k}(y_j)-σ(y_j)\|+\|σ(y_j)-σ(x)\|\leq 3ε$.
So the sequence converges uniformly on $I$
A: This really follows from the Pythagoras' theorem. We need to show $\sigma_n(t) \to \sigma(t)$ uniformly, or equivalently that $\sup_{[0,1]} \|\sigma_n(t) - \sigma(t)\| \to 0.$ Note by continuity of the function $(\sigma_n -\sigma)$ on $[0,1]$ for each $n,$ there's a point $t_n \in [0,1]$ where the said $\sup$ is attained. Join the line segment $\sigma_n(t_n)$ to $\sigma_n(0) = a,$ and similarly $\sigma_n(t_n) $ to $\sigma_n(1) = b.$ 
It follows by the mean value theorem that $\|\sigma_n(t) - \sigma_n(0)\| \leq L_n t,$ and $\|\sigma_n(t) - \sigma_n(1) \| \leq L_n(1-t),$ and similarly $\|\sigma(t) - \sigma(0)\| = Lt, $ and $\| \sigma(t) - \sigma(1) \| = L(1-t).$ 
Then by the pythagoras' theorem,
$$
\|\sigma_n(t) - \sigma(t)\|^2 \leq (L_n^2 - L^2)t^2, 
$$
and 
$$
\|\sigma_n(t) - \sigma(t) \|^2 \leq (L_n^2 - L^2)(1-t)^2.
$$
Multiplying the two and noticing $t(1-t) \leq 4^{-1},$ we are done. 
