By multiplying (4) and (5): \(K^2=
r\cdot r_a\cdot s(s-a)\)
By propositions
#13,
#14,
#15 triangles DFC and CGE are
similar (Similarity AA)
Therefore
\(\dfrac{r}{s-b}=\dfrac{s-c}{r_a}\implies
r\cdot r_a=(s-b)(s-c)\)
Substituting in (6):
\(K^2= s(s-a)(s-b)(s-c)\).
Therefore \(K=\sqrt {s(s-a)(s-b)(s-c)}
\text{ Q.E.D.}\)