Maximizing and minimizing functions in the Poincaré Half-Plane Model 
I've been studying hyperbolic isoperimetric problem and got stuck on the following situation: I have to prove that $P(x)$ has a unique minimum point at $x=0$ and $A(x)$ has a unique maximum point at $x=0$. The functions are defined:
$P(x) = arccosh(\cosh(c) . \cosh(d+x)) + arccosh(\cosh(c) . \cosh(d-x))$
$A(x) = 2arctan(\tanh(c) . \tanh(d+x)) + 2arctan ( \tanh(c) . \tanh(d-x))$
Where $c$ is the ray of the equidistant line $s$, $d$ is the lenght from $A$ to $O$ and $x$ is the lenght of a segment from an arbitrary point of the line $r$ to the point $O$.
My goal is to show that the triangle $ABG$ must be isosceles, but I'm stuck with this otimization problem. I tried differentiating, but the expressions got big and I'm lost. 
Thanks.
EDIT FOR BOUNTY:
The hyperbolic context can be ignored, then $c$ and $d$ are just constants. Just show that P(x) OR A(x) have a unique critical point and you get the bounty.
 A: $$\displaystyle \frac{d\, acosh(x)}{dx} = {{1}\over{\sqrt{x^2-1}}}$$
Taking $c$ and $d$ as constants.
$$\frac{dP(x)}{dx} = {{\cosh(c)\,\sinh \left(x+d\right)}\over{\sqrt{\cosh ^2(c)\,\cosh ^2
   \left(x+d\right)-1}}}+{{\cosh(c)\,\sinh \left(x-d\right)}\over{\sqrt{
   \cosh ^2(c)\,\cosh ^2\left(x-d\right)-1}}}$$
Note $\cosh(d-x) = \cosh(x-d)$.
$\cosh(x)$ is an even function and $\sinh(x)$ is an odd function so $x = 0$ satisfies $\displaystyle \frac{dP(x)}{dx} = 0$
but we need to prove its the only minimum.
For $\displaystyle \frac{dP(x)}{dx} = 0$ , cancel $\cosh(c)$ and multiply the denominators out:
$$\sinh (x+d) \sqrt{\cosh ^2(c)\,\cosh ^2\left(x-d\right)-1} = - \sinh (x-d) \sqrt{\cosh ^2(c)\,\cosh ^2(x+d)-1} $$
$$\sinh^2 (x+d) (\cosh ^2(c)\,\cosh ^2\left(x-d\right)-1) = \sinh^2 (x-d) (\cosh ^2(c)\,\cosh ^2(x+d)-1) $$
Using hyperbolic trig identities for $\sinh(x\pm y)$ and $\cosh(x\pm y):$
$\left(\cosh d\,\sinh x+\sinh d\,\cosh x\right)^2\,\left(\cosh ^2c\,
\left(\cosh d\,\cosh x-\sinh d\,\sinh x\right)^2-1\right)=\left(
\cosh d\,\sinh x-\sinh d\,\cosh x\right)^2\,\left(\cosh ^2c\,\left(
\sinh d\,\sinh x+\cosh d\,\cosh x\right)^2-1\right)$
Expanding:
$4\,\cosh ^2c\,\cosh d\,\sinh ^3d\,\cosh x\,\sinh ^3x-4\,\cosh ^2c\,
\cosh ^3d\,\sinh d\,\cosh x\,\sinh ^3x-4\,\cosh ^2c\,\cosh d\,\sinh 
^3d\,\cosh ^3x\,\sinh x+4\,\cosh ^2c\,\cosh ^3d\,\sinh d\,\cosh ^3x
\,\sinh x-4\,\cosh d\,\sinh d\,\cosh x\,\sinh x = 0$
$$ - $$
Divide out $4 \cosh(x) \sinh(x) \cosh(d) \sinh(d)$ , only $x = 0$ or $d = 0$ solves this factor:
$$\cosh ^2c\,\sinh ^2d\,\sinh ^2x-\cosh ^2c\,\cosh ^2d\,\sinh ^2x-
\cosh ^2c\,\sinh ^2d\,\cosh ^2x+\cosh ^2c\,\cosh ^2d\,\cosh ^2x-1 = 0$$
Collect terms around $\sinh^2 x$ and $\cosh^2 x$:
$$\cosh ^2c\,\sinh ^2x \, (\sinh ^2d\ -\cosh ^2d) +
\cosh ^2c\,\cosh ^2x \, (\cosh ^2d -\sinh ^2d) -1 = 0$$
$$- \cosh ^2c\,\sinh ^2x  + \cosh ^2c\,\cosh ^2x  -1 = 0$$
$$\cosh ^2c\, (-\sinh ^2x  + \cosh ^2x) = 1 $$
$$\cosh ^2c  = 1 $$
The functions of $x$ have reduced out.
Only $c = 0$ solves this factor but $c$ is  considered to be a constant.
$\displaystyle \lim_{x\to\infty} P(x) \to \infty$.
Note: let $\cosh(a) = \cosh(-a) = b$ so $arccosh(b) = a \, or \, -a$.
Since $P(x)$ is the length of the perimeter of ABG take the positive branch of $arccosh$ then $\displaystyle \lim_{x\to \, -\infty} P(x) \to \infty$
$x = 0$ is the only critical point $\left(  \frac{dP(x)}{dx} = 0 \right)$  , $P(0)$ is finite and both sides of it approach $+\infty$ which means $P(0)$ is a minimum.
Note: I used Maxima http://maxima.sourceforge.net/ to do most but not all of the algebra, for correctness.
A: The case $c = 0$ is trivial. I will focus on $c \neq 0$. Set $u = f.g$ where $f = \operatorname{arccosh},\; g = r \cosh$ and $r = \cosh(c).$ Since $P = u(d+x)+u(d-x)$ is even, it has a unique minimum at $0$ if $u$ is strictly concave upward. It suffices to show that $u'' > 0$ everywhere.
$$ \begin{align}
f' &= (-1+x)^{-1/2} (1+x)^{-1/2} \\
f'' &= -\frac12 (-1+x)^{-3/2} (1+x)^{-1/2} -\frac12 (-1+x)^{-1/2} (1+x)^{-3/2} \\&= -\frac12 \left(\frac1{-1+x} + \frac1{1+x}\right) f'\\
g' &= r \sinh \text{ so } (g')^2 = -r^2 + g^2\\
g'' &= r \cosh = g\\
u' &= f'.g\;\; g'\\
u'' &= f''.g \;\;g' g' + f'.g \;\;g'' = f'.g\; \left(-\frac12 \left(\frac1{-1+g} + \frac1{1+g}\right) (-r^2 + g^2) + g\right) \\&= f'.g \;\;g\; \left(1 - \frac{-r^2 + g^2}{-1 + g^2}\right) > 0 \text{ because }f'.g > 0\text{ and } g \geq r > 1 \text{ since } c \neq 0.
\end{align} $$
