圓D的半徑=\frac{a^2 (2 b-c)+(b-c)^2 (2 b+c)}{16 S}

A = (0.12813953488371593, 2.4636903304773567)
B = (-0.4090697674418667, 1.0132252141982836)
f = Line(A, B)
C = (1.9853488372093016, 0.9978763769889812)
g = Line(B, C)
h = Line(C, A)
i, j = AngularBisector(f, g)
k, l = AngularBisector(h, g)
I = Intersect(j, l)
P = Intersect(i, k)
m = OrthogonalLine(I, h)
n = OrthogonalLine(P, h)
E = Intersect(h, m)
cc = Circle(I, E)
F = Intersect(h, n)
d = Circle(P, F)
e = Circle(A, (Distance[A, E] * Distance[A, F] / Distance[A, C]))
C' = Intersect(e, h, 1)
D_{1} = Point(d)
p = Circle(C', C, D_{1})
G_{1} = Intersect(d, p, 2)
q = Line(D_{1}, G_{1})
H_{1} = Intersect(h, q)
r = Line(H_{1}, P)
h' = Mirror(h, r)
H = Intersect(d, h', 2)
s = Circle(C', H, C)
BC = Distance(B, C)
a: BC
CA = Distance(C, A)
b: CA
AB = Distance(A, B)
c: AB
IE = Distance(I, E)
S = Area(A, B, C)
o: ((a^(2) * ((2 * b) - c)) + ((b - c)^(2) * ((2 * b) + c))) / ((16 * S))
t = Radius(s)
D = Center(s)