Let $U$ be a domain in the plane and denote by $\mu_{x_0,U}$ the distribution of a Brownian motion starting at $x_0\in U$ by the time it hits the boundary of $U$.
Why is it that for $U=\mathbb{H}$ the upper half plane and any $x_0\in\mathbb{H}$, the exit distribution $\mu_{x_0,\mathbb{H}}$ is a Cauchy distribution on the horizontal axis (centered at the first component of $x_0$, and with scale parameter equal to the second component)? Note that the Cauchy distribution on the horizontal axis is the distribution of the intersections between the horizontal axis and random straight lines through the starting point with uniformly random angle.
Obviously this uniform angle property also holds for the exiti distribution from $U=\mathbb{D}$ the unit disk when $x_0=0$ (in this case the exit distribution is just the Hausdorff measure on the circle).
The question title was written in an attention gathering fashion though: I know this uniform-angle propery can't be true for all domains, but rather I'd like to ask what makes it true for the two special cases above and what makes it false for other domains.
EDIT: In a lame sense, the answer is "it's true for the half plane because the Cayley transform preserves BM up to time reparametrization and takes the horizontal axis to the circle and has derivative proportional to $1/(z+i)^2$, the absolute value of which is the Cauchy distribution for real $z$". This isn't very revealing to what's going on though, at least to me. I still don't have a clear picture of whether the uniform angle property maybe is true for all, e.g., convex domains if only formulated well enough (e.g. first pick a random line, then simulate a one dimensional Brownian motion hitting the boundary intersected with that line)