abc_4AR
Author: Andreas Brinken

Tipp 2

Tipp 1

Tipp 4

Tipp 3

A= \frac{ 1 }{2 } \cdot a \cdot h_a= \frac{ 1 }{2 } \cdot b \cdot h_b= \frac{ 1 }{2 } \cdot c \cdot h_c

Umfangswinkel-Mittelpunktswinkel-Satz

Mit  \;  h_c=  \frac{2 \cdot A}{c}  \; folgt:  \; 
 \frac{ R }{ \frac{ 1 }{2 }  \cdot a } = \frac{ b }{ \frac{2 \cdot A}{c}} 
 \; also:  \; 4\cdot A\cdot R=a\cdot b\cdot c

Der Beweis ''bricht zusammen'', wenn das Dreieck bei A 
einen stumpfen Winkel besitzt! Ist er zu retten?

Ähnlichkeit ausnutzen (www)

AA1 = (-2.9859091495799377, 6.515108004416202)
EE1: (x(AA1) + 1, y(AA1))
AA2: (x(AA1), y(AA1) - 0.4)
AA3: (x(AA1) + 4, y(AA1))
AA4: (x(AA3), y(AA3) - 0.4)
EE3: (x(AA3) + 1, y(AA1))
EE4: (x(AA3) + 1, y(AA1) - 0.4)
EE2: (x(AA1) + 1, y(AA1) - 0.4)
ss1 = Segment(AA1, EE1)
ss2 = Segment(AA2, EE2)
ss5 = Segment(AA3, EE3)
ss6 = Segment(AA4, EE4)
MM1 = Point(ss1)
MM2 = Point(ss2)
MM3 = Point(ss5)
MM4 = Point(ss6)
MP1 = Midpoint(ss1)
MP2 = Midpoint(ss2)
MP3 = Midpoint(ss5)
MP4 = Midpoint(ss6)
aa1 = Segment(AA1, MP1)
aa2 = Segment(AA2, MP2)
aa5 = Segment(AA3, MP3)
aa6 = Segment(AA4, MP4)
bb1 = Segment(MP1, EE1)
bb2 = Segment(MP2, EE2)
bb5 = Segment(MP3, EE3)
bb6 = Segment(MP4, EE4)
Anker1: (x(AA1), y(AA1) - 2)
A = (-2.785631815171581, -0.38333351409387223)
B = (2.3770728051327326, -3.0982040471849337)
C = (2.2880606565067962, 4.000514805733498)
poly1, c, a, b = Polygon(A, B, C)
d = Circle(A, B, C)
e = LineBisector(b)
f = LineBisector(a)
U = Intersect(e, f)
R = Segment(U, B)
j = OrthogonalLine(U, a)
E = Intersect(a, j)
α = Angle(B, U, E)
k = Line(A, B)
l = OrthogonalLine(C, k)
F = Intersect(k, l)
m = Line(C, B)
n = OrthogonalLine(A, m)
G = Intersect(m, n)
β = Angle(B, A, C)
h_c = Segment(C, F)
q = Segment(A, G)
γ = Angle(C, F, A)
δ = Angle(U, E, B)
r = Segment(U, E)
s = Line(A, C)