In an isosceles triangle $$ABC$$, with $$\angle ACB = 90^\circ$$, the point $$D$$ is on the side $$BC$$ such that $$\angle ADC = 75^\circ$$. If the area of triangle $$ADC$$ is 81, what is the length of segment $$BD$$?

Put $$AC = BC = s$$, $$CD = t$$ and $$BD = x = s - t$$, so the given area gives $$\tfrac{1}{2}st = 81$$, that is $$st = 162$$. Drawing $$DF \perp AB$$, triangle $$BDF$$ is a $$45^\circ$$ right triangle so $$BD^2 = 2DF^2$$, while $$\angle DAB = 30^\circ$$ gives $$AD = 2DF$$, hence $$AD^2 = 2BD^2 = 2x^2$$. Pythagoras in triangle $$ACD$$ now gives $$2x^2 = s^2 + t^2 = (s - t)^2 + 2st = x^2 + 324$$, so $$x^2 = 324$$ and $$BD = 18$$.