Proof of Schur's Theorem for Convex Plane Curves by Guggenheimer I'm reading Differential Geometry by Heinrich W. Guggenheimer and I have a doubt about the proof of Schur's Theorem for Convex Plane Curves on page 31. I will put the theorem and the proof here before I say what are my doubts.

$\textbf{Theorem 2-19.}$ Given two curves $f, g$ with continuous tangents and piecewise continuous curvature, both of length $L$. If $|k_f(s)| \geq |k_g(s)|$, $k_f(s) \neq k_g(s)$, the chord subtended by $g$ is bigger than that subtended by $f$.
$\textbf{Proof.}$
We place both arcs in the lower half plane $x_2 \leq 0$ with endpoints on the $x_1$ axis, $x_{2f}(0) = x_{2f}(L) = x_{2g}(0) = x_{2g}(L) = 0$ and such that $x_{1f} (L) > x_{1f} (0)$, $x_{1g} (L) > x_{1g} (0)$. In this case, both curvatures are non-negative. The chords subtended are $d_f = x_{1f}(L) - x_{1f}(0) = \int_0^L \cos \theta_f (s) ds$, $d_g = x_{1g}(L) - x_{1g}(0) = \int_0^L \cos \theta_g (s) ds$



Let $s'$ be the value of the arc length for which tangent to $f(s)$ is parallel to the $x_1$ axis (look at the picture above), then
$$\theta_f (s) = \int_{s'}^s k_f(\sigma) d \sigma$$
and, because of the convexity of the arc,
$$- \pi \leq \theta_f(s) \leq \pi$$
The angle
$$\theta_g^* (s) = \theta_g(s) - \theta_g(s') = \int_{s'}^s k_g(\sigma) d \sigma \leq \theta_f(s)$$
Hence
$$d_f = \int_0^L \cos \theta_f (s) ds \leq \int_0^L \cos \theta_g^* (s) ds = d_g \cos \theta_g(s') \leq d_g$$
and $d_f = d_g$ only if $k_f(s) = k_g(s)$ for all $s$. $\square$

My doubts are

*

*Why is true that $\int_0^L \cos \theta_g^* (s) ds = d_g \cos \theta_g(s')$?


*How the picture helps me to conclude the assertion of my first doubt?


*What is the geometric idea of the proof?
Thanks in advance!
 A: The picture. The picture is wrong. The angles in the picture don't agree with the geometric meaning of $\theta_f$ and $\theta^*_g(s)$. Also, if you look at the curve $f$, the angle $\theta_f(s)$ can lie in $[-3\pi/2, \pi/2]$, and not $[-\pi, \pi]$, as is mentioned in the proof.
Here is a correct picture

The geometric idea of the proof.
The first observation is that the angles are defined in terms of the curvature. Secondly, we also know that $d_f = x_{1f}(L) - x_{1f}(0) = \int_0^L \cos \theta_f(s)\,ds$  and
$d_g = x_{1g}(L) - x_{1g}(0) = \int_0^L \cos \theta_g(s)\,ds$. Intuitively, we are "adding up" all the $x_1$-components of the tangents along $f$ and $g$.
For these reasons, it is naturally to compare "the angle functions on $f$ and $g$".
But which angles should we compare with each other? $\theta_f$ and $\theta_g$?
No.
Note that $\theta_f$ and $\theta^*_g$ have the same geometric meaning for their respective curves $f$ and $g$. Both $\theta_f$ and $\theta^*_g$ measure the angle from the tangent at $s'$ to the tangent at $s$. That is why we obtain
$$
 k_g(s) \leq k_f(s) \Rightarrow \theta^*_g(s) \leq \theta_f(s).
$$
I think the picture wants to illustrate that we should compare $\theta_f$ and $\theta^*_g(s)$. But it would have been better if the point $g(s)$ was drawn on the arc between $g(s')$ and $g(L)$.
We have choosen $s'$ such that $T_f(s')$ lies in the $x_1$-direction. This makes it easy to get $d_f = \int_0^L \cos\theta_f(s)\,ds$. However, $T_g(s')$ is not necessarily horizontal, and thus we get $d_g\cos \theta_g(s') = \int_0^L \cos\theta^ *_g(s)\,ds$. (See the next paragraph.)
The equality. By the sum formula for cosine we get
$$
\begin{align*}
\int_0^L \cos \theta^*_{g}(s)\,ds 
 &= \cos\theta_g(s')\int_0^L \cos\theta_g(s)\,ds + \sin \theta_g(s')\int_0^L \sin\theta_g(s)\,ds \\
&= \cos \theta_g(s') \left(x_{1g}(L)-x_{1g}(0)\right) + \sin \theta_g(s') \left(x_{2g}(L)-x_{2g}(0)\right) \\
&= \cos \theta_g(s')\, d_g. 
\end{align*}
$$
You can also get this equality in a geometric way.
The integral $\int_0^L  \cos\theta^*_g(s)\,ds$ is, by definition of $\theta^*_g(s')$,  the length of the $T_g(s')$-component of the segment $x_{1g}(L)-x_{1g}(0)$. (The reasoning is similar to the equation $\int_0^L \cos\theta_g(s)\,ds = x_{1g}(L)-x_{1g}(0)$.) The angle marked with 'v' is $\theta_g(s')$, so this segment has length $d_g \cos\theta_g(s')$.
