The bivariate distribution formed by two independent normalized Gaussians is rotationally symmetric (think about the usual argument for evaluating the probability integral). The quotient of two random variables $X$ and $Y$ is the tangent of the angle between $(0,0)$ and $(X,Y)$ with the $x$-axis. If one has a rotationally symmetric distribution for $X$ and $Y$ (with no point mass at the origin) then $Y/X$ is a tangent of a uniformly distributed angle. This is the Cauchy distribution.
Added Your example with the Brownian motion states in effect that if $P$ is the first point that the motion hits the $x$-axis then the angle between the line from $P$ to the starting point and the $y$-axis is uniformly distributed between $-\pi$ and $\pi$. I can't see any reason why this should be so, but perhaps someone (unlike me) who actually knows something about Brownian motion might know why.