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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, $\alpha=\beta$.

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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 $a+b=c$.

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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 $$A=\frac{1}{2}4r^2\sin(\gamma)=2r^2\sin(\gamma)\le 2r^2$$

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