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.