A geometry problem by ritwik jain

If $\angle ACB = 2x^\circ$ , then ( $r$ is the inradius) $\frac{2r}{BC} = \tan x^\circ$

Since $BH$ is perpendicular to $AC$ , $\angle HBD = (90 - 2x)^\circ$ , and $\frac{1}{\tan2x^\circ} \; = \; \tan(90 - 2x)^\circ \; = \; \frac{4r}{BC} \; = \; 2\tan x^\circ$ Thus $2\tan x^\circ \tan2x^\circ = 1$ , so that $\tan x^\circ =\frac{1}{\sqrt{5}}$ , so $\tan2x^\circ = \tfrac12\sqrt{5}$ and hence $\frac{BC}{2AB} \; = \; \cos2x^\circ\; = \; \tfrac23$ so that $\frac{AB}{BC} = \boxed{\tfrac34}$ .