Perimeter change when moving one point of a triangle along a circle I am interested in the perimeter $P(\Delta')$ of a modified version of a triangle $\Delta = (A,B,C)$. I chose point $B$ to be the point I want to shift towards the inside (or on the boundary) of $\Delta$. I chose a distance $r<h$ where $h$ is the height of the triangle. $B$ can be moved to an arbitrary point $E$ which is of distance $r$ to $B$ and lays inside or on the boundary of $\Delta$. This means that $E$ must also lay on the circle of radius $r$ with the center point $B$. I denote $\Delta'=(A,E,C)$. What is the perimeter $P(\Delta')$ of the new triangle given the original side lengths, $r$ (and probably some representation of the arc length to E)? In what range does $P(\Delta)-P(\Delta')$ lie in?
I attached a small figure which visualizes both triangles.

Edit
Based on the comments, I added some notation:
Assuming $B=(0,0)$, then we have $E=(x,y)$ with $x=\lambda_A x_A + \lambda_C x_C$ and $y=\lambda_A y_A + \lambda_C y_C$ ($\lambda_A+\lambda_C = 1$).
This gives us the following set of equations:
$$r=\sqrt{(0-x)^2 + (0-y)^2} = \sqrt{(\lambda_A x_A + \lambda_C x_C)^2 + (\lambda_A y_A + \lambda_C y_C)^2}$$
$$|AE| = \sqrt{(x-x_A)^2+(y-y_A)^2}$$
$$|CE| = \sqrt{(x-x_C)^2+(y-y_C)^2}$$
Moreover, we have:
$$|AB| = \sqrt{x_A^2 + y_A^2}$$
$$|CB| = \sqrt{x_C^2 + y_C^2}$$
The perimeter change is in fact the difference $|AB|+ |CB| - |AE| - |CE|$. I would like to show that $r \leq |AB|+ |CB| - |AE| - |CE|$. I struggle with the proof because I still have six variables $x_A,x_C,y_A,y_C,\lambda_A$ and $\lambda_C$.
Background
This question arises from a computer science problem called lawn mowing/milling, where you are given a cutter (in my case a circular shape) and a Polygon to cover. I noticed that given a tour of the circular cutter one can change the cutter from a circular shape with a radius $s$ to a square with side length $2s$. Afterward, the tour can be shortened a bit. I have already proven a specific length $r$ that I can move the points of the tour "inwards" to shorten the tour. My goal is to find out how much shorter the final tour gets when I modify it. In my example, $A-B-C$ are part of the tour and $B$ is a point which can be moved inward by $r$ to a point $E$. My tour will later be $A-E-C$. I want to find out how much the tour is shortened after my transformation and I hope that it will be shortened by at least $r$.
 A: (Expanding a comment.)

I would like to show that
$$r \leq |AB|+|CB|−|AE|−|CE| \tag1$$

Without additional conditions, $(1)$ may not be true. Writing the inequality as
$$|AE|+|CE|\leq |AB|+|CB|-r \tag{1'}$$ leads us to consider the locus of points $P$ satisfying the corresponding equality:
$$|AP|+|CP|=|AB|+|CB|-r \tag2$$
This is an ellipse with foci $A$ and $C$, with major radius $\frac12(|AB|+|CB|-r)$. This ellipse separates the points with a larger sum from those with a smaller sum. The described points $E$, then, must lie on the arc where $\bigcirc B$, with radius $r$, overlaps this ellipse; depending upon $r$, this arc may-or-may-not include all (or even any!) points of the arc overlapping the triangle itself.
For instance, this first figure shows a situation with points $E_-$, $E_0$, $E_+$ on $\bigcirc B$ and inside $\triangle ABC$, but respectively inside, on, and outside the ellipse:

That is, we have
$$\begin{align}
|AE_-|+|CE_-| < |AB|+|BC|-r \quad\to\quad r < |AB|+|BC|-|AE_-|-|CE_-|\\
|AE_0\;|+|CE_0\,| = |AB|+|BC|-r \quad\to\quad r = |AB|+|BC|-|AE_0\,|-|CE_0\;|\\
|AE_+|+|CE_+| > |AB|+|BC|-r \quad\to\quad r > |AB|+|BC|-|AE_+|-|CE_+|
\end{align} \tag3$$
so that only $E_-$ and $E_0$ satisfy $(1)$.
With appropriate conditions, we can have a circumstance where the circle-triangle overlap is contained within the ellipse (which seems to be the case OP envisions):

On the other hand, we can have a case where the circle doesn't meet the ellipse at all, so all points on the circle are in the $E_+$ family that don't satisfy $(1)$:


For a given triangle, the threshold $r$-values that keep the circular arc within the ellipse can be determined by intersecting the circle with the ellipse. Unfortunately, conic-conic intersections like this lead to quartic equations, which are exceedingly messy to solve in general.
Unless/until OP provides additional constraints or context, this is about as far as this investigation will go.
A: This answer deals with the minimal perimeter issue.

Using slightly different notation, we have
a known $\triangle ABC$ and given radius $r_a$
of the circle $\mathcal{A}(A,r_a)$.
The circle intersects with $AB$, $AC$ at $D,\ E$, respectively.
For the point $F$ along the interior arc $DE$,
and the point $G=AF\cap BC$,
the minimal perimeter of $\triangle FBC$
is achieved when $F$ is the tangential point,
where the ellipse focused at $B,C$ touches
the circle $\mathcal{A}$, which also means
that $\angle GFB=\angle CFG=\theta$.
Let $|BF|=u,\ |CF|=v$, $|BG|=at,\ |CG|=a(1-t)$
for some $t\in(0,1)$.
Then using the cosine rule for $\triangle ABG$,
Stewart’s theorem
for $\triangle FBC$ and the properties of
the bisector of $\angle CFB$, we can express $t$ in terms of
the side lengths $a,b,c$ of $\triangle ABC$ and the radius $r_a$
as the root of the quartic
\begin{align} 
q_4\,t^4+q_3\,t^3+q_2\,t^2+q_1\,t+q_0&=0
\tag{1}\label{1}
,
\end{align}
where
\begin{align} 
q_4 &= (b^2-c^2)^2-4\,r_a^2\,a^2,
\\
q_3 &= 4\,c^2\,(b^2-c^2)-4\,r_a^2\,(b^2-2\,a^2-c^2),
\\
q_2 &= r_a^2\,(4\,b^2-8\,c^2-5\,a^2)+2\,c^2\,(3\,c^2-b^2),
\\
q_1 &= r_a^2\,(a^2-b^2+5\,c^2)-4\,c^4,
\\
q_0 &= c^2\,(c^2-r_a^2)
\tag{2}\label{2}
,
\end{align}
if the vertices $A,B,C$ in counter-clockwise orientation.
The real roots of \eqref{1} need to be checked
for which one gives the actual minimum.
For the attached image, I've got all four real roots,
two of them were greater than $1$, one gives the minimal perimeter,
and, unfortunately, another one was also in a valid range,
but the corresponding perimeter was not minimal.
Also, the curved red, green and blue lines,
shown in the image,
emanating from the vertices $A,B$ and $C$, respectively,
illustrate the corresponding locus
of the points of minimal perimeter
for possible values of $r_a,\ r_b$ and $r_c$
for the circles
$\mathcal{A}(A,r_a)$,
$\mathcal{B}(B,r_b)$
and $\mathcal{C}(C,r_c)$.
Naturally, if the angles of $\triangle ABC$ are less than $120^\circ$,
all this curves intersect at the
Fermat–Torricelli point
$T$, which provides the minimum of $|TA|+|TB|+|TC|$.

Edit
On the other hand, it is much simpler to find
the corresponding radius $r_a$ given some $t\in(0,1)$:
\begin{align}
r_a
&=
\left|
\frac{t^2 b^2-(1-t)^2\,c^2}
{(1-2t)\sqrt{(c^2-a^2\,t)(1-t)+b^2\,t}}
\right|
.
\end{align}
Note that $t=\tfrac12$ is a special (and simpler) case,
which must agree only with $b=c$,
and such condition is better considered separately.
For $b=c$, the equation \eqref{1}
factors out as
\begin{align}
(1-2t)^2(b^4-r_a^2(b^2-a^2\,t\,(1-t)))
&=0
,
\end{align}
and the proper root is $t=\tfrac12$,
so, for any valid value of $r_a$,
$G$ must be the midpoint of $BC$,
to get the minimal perimeter of $\triangle FBC$,
as expected for the isosceles $b=c$ case.

Edit 2
Actually, the valid range of $t$ would be a subset of either
$(0,\tfrac12)$ or $(\tfrac12,1)$, depending of
the side lengths $b$ and $c$.
A: We can find perimeter change as a function of parameter $ \theta$
Let the circle be centered at (0,0) and radius r.
Let two points on a vertical  line be $ ( h,-p),(h,q) $
slant line lengths sum for $\theta_1,$ $ L1=$
$$ = \sqrt { r\cos \theta_1 -h)^2} + \sqrt{( r \sin \theta_1 +p)^2}$$
$$ + \sqrt { r\cos \theta_1 -h)^2} + \sqrt{( r \sin \theta_1-q)^2}$$
For $\theta_2$, $L_2=$
$$ = \sqrt { (r\cos \theta_2 -h)^2} + \sqrt{( r \sin \theta_2 +p)^2}$$
$$ + \sqrt { (r\cos \theta_2 -h)^2} + \sqrt{( r \sin \theta_2-q)^2}$$

The perimeter change is  difference  $ L_2- L_1$ where $L_2$ pair is marked in red. It depends on two $\theta_ s$ , the constants are  $( h,p,q,r).$
