Geometry problem using circular arcs The construction for the problem is as follows:

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*Given some circular arc $A$ centred at $C$ with an angle $\theta \geq 180^{\circ}$ and endpoints $a,b$, take some arbitrary point $t$ inside the region bounded by the arc and the segment $ab$.


*Construct two arcs, $A_1, A_2$ within $A$ by taking the intersection of the perpendicular bisector of $at$ / $tb$ and the segment $aC$ / $Cb$. The endpoints of the arcs are at $a, t$ and $t, b$ respectively, and the centre at the aforementioned intersection point.


*The problem is to show that $\mathbf{A_1 + A_2 \leq A}$ for any choice of $\mathbf{t}$.
Construction:

Some initial points are that clearly if $t$ is on $A$, then $A_1, A_2$ just make up the entire arc, and then if $t$ is on the segment $ab$ then you can show using similar triangles that $A_1 + A_2 = A$ in this case. While this problem seems intuitively true, I haven't been able to come up with any way of proving it definitively.

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The picture above illustrates how $A_1 + A_2$ depends on the position of $T$. For every $T$ there is a point on the vertical axis with the same sum of the arcs. Therefore we can consider a unit circle with center $C = (0,c)$ that intersects the horizontal axis at $(0,\pm x)$ where $x=\sqrt{1-c^2}$ and place $T$ at $(0,y)$
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The length of the arcs is then given by $A = 2\arccos{(x)} + \pi$ and $A_1 = A_2 = r \cdot \alpha$ where
$$\begin{matrix} \displaystyle r = \sqrt{a^2 + \left(y - c \left(\frac{a}{x} + 1 \right) \right)^2} \\ \displaystyle \alpha = \arccos{\left( \frac{a + x}{1 - c^2} \, \frac{a + cx(c-y)}{r^2} \right)} \end{matrix} \qquad a = \frac{x}{2} \, \frac{(c-y)^2 - 1}{1 + c(y-c)}$$
So your problem becomes just algebra $$\arccos{(x)} + \frac{\pi}{2} > r \cdot \begin{cases} \alpha \\ (2 \pi - \alpha)\end{cases} \begin{matrix} \text{if} \; 0 < y - c < 1 \\ \text{if} \; 0 < y < c \end{matrix}$$
