Let $$ABC$$ be a triangle with $$AB = AC$$. Let $$D$$ be a point on the segment $$BC$$ such that $$BD = 48\frac{1}{61}$$ and $$DC = 61$$. Let $$E$$ be a point on $$AD$$ such that $$CE$$ is perpendicular to $$AD$$ and $$DE = 11$$. Find $$AE$$.

Let $$M$$ be the midpoint of $$BC$$, so $$AM$$ is perpendicular to $$BC$$. Here $$BC = \frac{2929}{61} + 61 = \frac{6650}{61}$$, so $$BM = \frac{3325}{61}$$ and $$DM = BM - BD = \frac{3325 - 2929}{61} = \frac{396}{61}$$. The right triangles $$ADM$$ and $$CDE$$ share the angle at $$D$$, hence they are similar and $$\frac{DM}{DA} = \frac{DE}{DC}$$, giving $$AD = \frac{396}{61} \times \frac{61}{11} = 36$$. Therefore $$AE = AD - DE = 36 - 11 = 25$$.