lwma5
Author: Andreas Brinken

\mbox{Zu zeigen:  B' liegt auf der Geraden DG}

Beweisidee

Aufgabe

Schiebe den Regler für α nach rechts!

Trick

\mbox {Spiegle das Dreieck } \DELTA \mbox {FBC an C.}

Kongruenzsatz
            q. e. d.

Beweisende

sws

\mbox{Zeige: } \overline{GF}=\overline {FB}

Radius = 0.49 (slider from 0 to 1, step 0.01)
Basis = 5.5 (slider from 0 to 8, step 0.1)
A = (-5.115238282114596, -3.099047135850271)
B = Point(Circle[A, Basis])
a = Segment(A, B)
C_1 = Midpoint(a)
c = Circle(C_1, B)
b_2 = LineBisector(a)
C = Intersect(c, b_2, 2.0)
Vieleck1, d, a_1, b = Polygon(A, B, C)
γ = Angle(A, C, B)
e = Circle(C, Radius * b)
D = Intersect(e, b, 1.0)
E = Intersect(e, a_1, 1.0)
AE = Line(A, E)
DG = OrthogonalLine(D, AE)
CF = OrthogonalLine(C, AE)
G = Intersect(DG, d)
F = Intersect(CF, d)
H = Intersect(AE, DG)
I = Intersect(AE, CF)
β = Angle(E, I, C)
γ_1 = Angle(I, H, D)
δ, ε, ζ = Angle(Vieleck1)
B' = Mirror(B, C)
C' = Mirror(C, C)
Vieleck2, c'_1, CE, e_1 = Polygon(D, E, C')
CB' = Segment(C', B')
F' = Mirror(F, C')
B'F' = Segment(B', F')
CF' = Segment(F', C)
FG = Segment(G, F)
FB = Segment(F, B)
α = 0 rad (slider from 0 to 1.5707963267948966, step 0.017453292519943295)
Vieleck3, c_1, a_2, e_2 = Polygon(A, E, C)
A' = Rotate(A, -1.0 * α, C)
E' = Rotate(E, -1.0 * α, C)
C'_1 = Rotate(C, -1.0 * α, C)
Vieleck3', c'_2, CD, e' = Polygon(A', E', C'_1)
reDrehwinkel = Angle(A', C'_1, E')
idee1 = false
aufgabe = false
idee2 = false
ende1 = false
AC = Segment(A, C)
BC = Segment(B, C)
B'C = Segment(B', C)
κ = Angle(B', C, A)
A'C = Segment(A', C)
CAE, c_3, a_3, e_3 = Polygon(A, E, C)
CB'D, c_4, b', d_1 = Polygon(B', D, C)
BCF, b_1, c_2, f = Polygon(C, F, B)
B'CF', b'_1, c_5, f' = Polygon(C, F', B')
η = Angle(AE, CF)
Vieleck4, g, f_1, f'_1, b'_2 = Polygon(G, F, F', B')