Extremal Area and side length of a triangle in a circle

Given a Circle $$B_r$$ and a triangle $$\triangle ABC$$ in it with two fixed (immovable) points $$A$$ and $$B$$. The third point $$C$$ can be moved on the circle. I want to prove the following equivalence: (for a visualization see picture below)

The area of the triangle $$\triangle ABC$$ is extremal (with regard to the position of $$C$$) if and only if the sum of the lengths $$|AC|+|CB|$$ is extremal (with regard to the position of $$C$$).

I know already that this is the case if the Point $$C$$ is chosen to be the point at which the Tangent $$T_c B_r$$ is parallel to $$|AB|$$ or, equally, $$\alpha=\beta$$.

Let $$a=BC$$, $$b=AC$$, $$c=AB$$, $$\gamma=\angle ACB$$. Combining $$2ab=(a+b)^2-(a^2+b^2)$$ with $$a^2+b^2=c^2+2ab\cos\gamma$$ we get: $$ab={(a+b)^2-c^2\over2(1+\cos\gamma)} \quad\text{and}\quad area_{ABC}={1\over2}ab\sin\gamma={(a+b)^2-c^2\over4(1+\cos\gamma)}\sin\gamma.$$ As long as $$C$$ lies on one of the arcs $$AB$$ the value of $$\gamma$$ is fixed, as is fixed $$c$$. Hence the area depends only on $$(a+b)$$ and it is extremal if $$(a+b)$$ is.
Notice however that $$\gamma$$ changes to $$\pi-\gamma$$ when $$C$$ passes from one arc to the other: $$\sin\gamma$$ doesn't change, while $$\cos\gamma$$ changes its sign. Hence there are two local maxima, while the minimum (area$$\ =0$$) is attained in the degenerate case $$a+b=c$$.
We can use the formula $$A=\frac{1}{2}ab\sin(\gamma)$$ with $$\sin(\alpha)=\frac{a}{2r},\sin(\beta)=\frac{b}{2r}$$ we get $$A=\frac{1}{2}4r^2\sin(\gamma)=2r^2\sin(\gamma)\le 2r^2$$