Let $$ABC$$ be a triangle with $$AB = 5$$, $$AC = 4$$, $$BC = 6$$. The internal angle bisector of $$C$$ intersects the side $$AB$$ at $$D$$. Points $$M$$ and $$N$$ are taken on sides $$BC$$ and $$AC$$, respectively, such that $$DM \parallel AC$$ and $$DN \parallel BC$$. If $$(MN)^2 = \frac{p}{q}$$ where $$p$$ and $$q$$ are relatively prime positive integers then what is the sum of the digits of $$|p-q|$$?

The bisector gives $$\frac{AD}{DB} = \frac{CA}{CB} = \frac{4}{6}$$, so $$AD = 2$$ and $$DB = 3$$, and $$DMCN$$ is a parallelogram whose diagonal $$CD$$ bisects angle $$C$$, making it a rhombus of side $$DM = 4 \times \frac{3}{5} = \frac{12}{5}$$. The bisector length satisfies $$CD^2 = CA \cdot CB - AD \cdot DB = 24 - 6 = 18$$, and for a rhombus $$\left(\frac{CD}{2}\right)^2 + \left(\frac{MN}{2}\right)^2 = \left(\frac{12}{5}\right)^2$$. This gives $$MN^2 = \frac{126}{25}$$, so $$|p-q| = 101$$ and its digit sum is 2.