Probability that 3 darts land in a same half of a dart board

Suppose rays are drawn from the center of the disk to the places where the first two darts landed. And suppose for example the angle between those two rays is $23^\circ.$ Say you rotate the ray to the first dart $23^\circ$ counterclockwise to get the ray to the second dart. If you go counterclockwise from that ray $180^\circ$ you will sweep out a part of the disk within which, if the third dart lands there, then all three will be on the same side of that line formed from that ray and the ray $180^\circ$ counterclockwise from it. But also, if you start with the ray to the second dart and go $180^\circ$ clockwise you will sweep out a region within which, if the thrid dart lands there, then all three will be in that half of the disk.

Thus you will fail to put all three on the same side of some line through the center only if the third dart lands in the region $23^\circ$ wide that is opposite the one bounded by the the two aforementioned rays. Thus the probability that they would be in a common half of the disk is $$ \frac{360^\circ-23^\circ}{360^\circ}. $$

All of that holds if the angle is $23^\circ.$ But the angle is uniformly distributed between $0^\circ$ and $180^\circ,$ so what you need is the expected value of $$ \frac{360^\circ - (\text{random angle})}{360^\circ}. $$ Since an angle uniformly distributed between $0^\circ$ and $180^\circ$ is on average $90^\circ,$ the random quantity above is on average $$ \frac{360^\circ-90^\circ}{360^\circ} = \frac 3 4. $$

Let us represent impact points by their polar coordinates $(r_k,\theta_k)$ on a unit radius disk. An essential preliminary remark is that we need only consider angles $\theta_k$ (see detailed explanations in the Edit below). Therefore, this question is geometricaly equivalent to the following one: being given a triangle with vertices on the unit circle, what is the probability that this triangle doesn't contain the origin ?

This issue has been treated here where several answers give $1/4$ for the probability of the complementary event (the triangle contains the origin). Therefore, the probability that the darts belong to a same half of the dartboard is $1-1/4=3/4$.

Remark: A somewhat connected issue can be found here.

Edit: (following a remark by @BlueRaja - Danny Pflughoeft) More precisely, on the geometrical side, saying that all impact points $I_k:=r_ke^{i\theta_k}$ are in the same half-plane is equivalent to say that all points on the unit circle $J_k:=e^{i\theta_k}$ are in the same half-plane.

[an even more precise reason is that "being in the same half-plane" is a property attached to the "convex hull" of all points $ae^{i\theta_1}+be^{i\theta_2}+ce^{i\theta_3}$ with positive values of $a,b,c$.]

On the probability side, using terms defined here, the new underlying σ-algebra is a quotient space of the initial one with a "canonical" transfer of the probability law.