In triangle $$ABC$$, let $$P$$ and $$R$$ be the feet of the perpendiculars from $$A$$ onto the external and internal bisectors of $$\angle ABC$$, respectively; and let $$Q$$ and $$S$$ be the feet of the perpendiculars from $$A$$ onto the internal and external bisectors of $$\angle ACB$$, respectively. If $$PQ = 7$$, $$QR = 6$$ and $$RS = 8$$, what is the area of triangle $$ABC$$?

Reflecting $$A$$ in each bisector lands on line $$BC$$, so $$P, Q, R, S$$ are the midpoints of $$A$$ with the points at signed positions $$-c$$, $$a-b$$, $$c$$ and $$a+b$$ on line $$BC$$ measured from $$B$$, and all four lie on the midline. Halving the distances gives $$a-b+c = 14$$, $$|b+c-a| = 12$$ and $$a+b-c = 16$$, whose solution is $$a = 15$$, $$b = 14$$, $$c = 13$$. By Heron's formula the area is $$\sqrt{21 \times 8 \times 7 \times 6} = 84$$.