If we rotate the coordinate system so that the ramp is "horizontal", then the free-fall acceleration will have two components; one downward given by a_y' = –g cosθ and one horizontal given by a_x' = g sinθ . In this frame, the initial velocity will have components given by v_y = v₀ cosθ and v_x = v₀ sinθ. 
The time for each bounce is given by

${\mathrm{t}}_{\mathrm{b}}=\frac{2{\mathrm{v}}_{\mathrm{y}}}{-{\mathrm{a}}_{\mathrm{y}}}=\frac{2{\mathrm{v}}_0}{\mathrm{g}}$

The horizontal displacement is given by

$\mathrm{\Delta x}={\mathrm{v}}_{\mathrm{x}}\mathrm{t}+0.5{\mathrm{a}}_{\mathrm{x}}{\mathrm{t}}^2$

Given these two equations, we can find     

${\mathrm{d}}_{12}={\mathrm{v}}_0\sin \mathrm{\theta }\left(\frac{2{\mathrm{v}}_0}{\mathrm{g}}\right)+0.5\mathrm{gsin}\mathrm{\theta }{\left(\frac{2{\mathrm{v}}_0}{\mathrm{g}}\right)}^2=\frac{4{\mathrm{v}}_0^2\sin \mathrm{\theta }}{\mathrm{g}}$

and 

${\mathrm{d}}_{13}={\mathrm{v}}_0\sin \mathrm{\theta }\left(\frac{4{\mathrm{v}}_0}{\mathrm{g}}\right)+0.5\mathrm{gsin}\mathrm{\theta }{\left(\frac{4{\mathrm{v}}_0}{\mathrm{g}}\right)}^2=\frac{12{\mathrm{v}}_0^2\sin \mathrm{\theta }}{\mathrm{g}}$

Since, d₂₃ = d₁₃ – d₁₂, the ratio

$\frac{{\mathrm{d}}_{12}}{{\mathrm{d}}_{23}}=\frac{1}{2}$