Date: 10 Únor 2014

\text{Ak bod }K \text{ je vnútorný bod trojuholníka }ABC\text{, tak platí: }S=\frac{AF}{BF}×\frac{BD}{DC}×\frac{CE}{EA}=1

Dôkaz

\text{Z podobnosti trojuholníkov dostávame:} 

 \frac{AF}{BF}=\frac{AH}{BC}

T_1

 \frac{CE}{EA}=\frac{BC}{AG}

T_2

T_3

  \frac{BD}{AG}=\frac{DK}{AK} ⇒AG=AK×\frac{BD}{DK}

  \frac{DC}{AH}=\frac{DK}{AK} ⇒AH=AK×\frac{DC}{DK}

T_4

 \text{Odkiaľ dostávame }\frac{AH}{AG}=\frac{BD}{DC}

 \text{Po dosadeni do vzťahu }S \text{ dostávame tvrdenie.}

o = true
A = (-0.3592055138686547, -1.244381518768267)
B = (1.6649272756993048, -1.5639814329105755)
C = (1.1921857360304722, -0.1124651561809246)
a = Segment(A, B)
b = Segment(B, C)
c = Segment(C, A)
d = Line(C, B)
e = Line(A, B)
f = Line(C, A)
troj, c_1, a_1, b_1 = Polygon(A, B, C)
K = PointIn(troj)
h = Line(B, K)
g = Line(A, K)
D = Intersect(g, d)
E = Intersect(h, f)
EA = Distance(E, A)
CE = Distance(C, E)
i = Line(C, K)
F = Intersect(i, e)
FB = Distance(F, B)
DC = Distance(D, C)
BD = Distance(B, D)
h2: BD / DC
AF = Distance(A, F)
h1: AF / FB
h3: CE / EA
h4: ((h1 * h2) * h3)
j = Segment(C, F)
k = Segment(B, E)
l = Segment(A, D)
m = Segment(C, E)
n = Segment(B, F)
p = Segment(C, D)
q = Segment(B, K)
r = Segment(C, K)
s = Segment(A, K)
t = Line(A, b)
G = Intersect(h, t)
H = Intersect(i, t)
d_1 = Segment(G, E)
e_1 = Segment(H, F)
f_1 = Segment(G, H)
mnohoúhelník1, f_2, a_2, h_2 = Polygon(A, H, F)
mnohoúhelník2, f_3, b_2, c_2 = Polygon(B, C, F)
u = false
mnohoúhelník3, g_2, a_3, e_2 = Polygon(A, E, G)
mnohoúhelník4, e_3, b_3, c_3 = Polygon(B, C, E)
v = false
mnohoúhelník5, k_1, a_4, g_3 = Polygon(A, G, K)
mnohoúhelník6, k_2, b_4, d_2 = Polygon(B, D, K)
w = false
mnohoúhelník7, k_3, c_4, d_3 = Polygon(C, D, K)
mnohoúhelník8, k_4, a_5, h_3 = Polygon(A, H, K)
h_1 = false