Among all triangles ABC with same perimeter and same base AB, which has bigger area? I proved that the isosceles triangle has bigger area, using Heron's formula.
But, I couldn't prove this fact:
Given two triangles ABC and ABC' with same perimeter, if $|\overline{AC}-\overline{BC}|<|\overline{AC'}-\overline{BC'}|$, then the area of ABC is bigger than ABC'.
Sketch of the solution:
I tried to show that $(p-b)(p-c)>(p-d)(p-e)$, where $b=\overline{BC}, c=\overline{AC}$ and $d=\overline{BC'}, e=\overline{AC'}$, to use Heron's formula, but I can't get anywhere from that. I also used the fact that $b+c=d+e$, since the two triangles have the same perimeter and the same side AB.
 A: The maximum area is attained by the isosceles triangle. A proof without words:

For the second part, exploit the convexity/concavity of hyperbolas (loci such that a difference of distances is constant) and ellipses (loci such that a sum of distances is constant).
A: As it was already noted, the third vertex, 
$C$ is located on an ellipse.

Let $|BC|=a,\ |AC|=b,\ |AB|=c$ be the sides of $\triangle ABC$,
and let $|OP|=u,\ |OQ|=v$ be 
the  semi-major and semi-minor axis of the ellipse, respectively.
Due to the symmetry, we can consider only the I quadrant of the ellipse.
Following the cosine rule,
\begin{align}
a^2&=b^2+c^2-2bc\cos\alpha
,\\
b^2&=a^2+c^2-2ac\cos\beta
,\\
2(b^2-a^2)&=2bc\cos\alpha-2ac\cos\beta
,\\
b-a&=\frac{c}{a+b}(b\cos\alpha-a\cos\beta)
\\
&=
\frac{c}{a+b}(|AH|-|BH|)
=
\frac{2\,c}{a+b}|OH|
=
\frac{2\,c}{a+b}x
.
\\
\text{Since }
\frac{2\,c}{a+b}&=\mathrm{const},
\quad
b-a\sim x
.
\end{align}  
Also, the area of $\triangle ABC$ is directly proportional to its height,
that is, $[ABC]\sim y$.
Expression for the point $C=(x,y)$ on the elliptic arc $PQ$
in terms of parameter $\theta\in(0,\tfrac\pi2)$ gives
\begin{align}
x(\theta)&=u\cos\theta
,\\
y(\theta)&=v\sin\theta
,\\
x'(\theta)&=-u\sin\theta<0\quad \forall\theta\in(0,\tfrac\pi2)
,\\
y'(\theta)&=v\cos\theta>0\quad \forall\theta\in(0,\tfrac\pi2)
,
\end{align}
so, smaller $x$ is equivalent to bigger $y$
and the smaller the side difference,
the bigger the area.
A: Well, I don't know if this is appropriate answer, but here it goes. $C$ and $C'$ must lie on a the same ellipse since we have:
$$AC+BC = AC'+BC'$$ 
Imagine this ellipse is with focuses $A$ and $B$ on x-axis. Since  $|\overline{AC}-\overline{BC}|<|\overline{AC'}-\overline{BC'}|$ the triangle is ''more'' isosceles then $ABC'$. So $ABC$ has bigger altitude on $AB$ then $ABC'$ does and thus the conclusion.
A: Using the Heron formula we obtain that the areas of $ABC$ and $ABC'$are respectively $S$ and $S'$. Suppose $H$ and $H'$are the heights relative to the $AB$ side. Write 
$$S= \frac{AB * H}{2} \,\,\,\ \textrm{and}\,\,\,\ S'= \frac{AB * H'}{2}.$$
Use the hypothesis and the fact that $AC'+BC'=AC+BC$
