\; \text{Dokážte, že bod } S_1=\overrightarrow{CO}∩k \text{ je stredom kružnice }ω\text{ opísanej }\triangle AOB.

\math{\pmb k} \text{  - kružnica opísaná }\triangle ABC

\math{\pmb O} \text{  - stred kružnice vpísanej }\triangle ABC

 \text{Postup} 

$\pmb γ$

$\pmb γ$

1.  \;  ∡ ACS_1≅∡ BCS_1; \; o_1 \text{ je os uhla }∡ ACB

2.  \; \text{ Odtiaľ pre oblúky: } \widehat{AS_1}≅\widehat{BS_1}  \; ∧ \; AS_1≅BS_1

$\pmb β$

3.  \; \text{ Vonkajší uhol }φ:∡S_1OB=β+γ } \triangle  \text{ ... }

γ

4.  \; \text{ Obvodový uhol } ∡ABS_1=γ

$\pmb β$

$\pmb φ$

5.  \; \text{ Zrejme } ∡OBS_1=β+γ=φ; \; o_2\text{ je os uhla}

6. \; \triangle BOS_1\text{ je rovnoramenný, odkiaľ } BS_1≅OS_1 ( ... ≅AS_1)

7. \; \text{Body }A,O,B∈ω(S_1,r=S_1O)

A = (2.0091622113398437, 2.7195657184772535)
B = (9.001239039652052, 2.8336738940659165)
C = (2.5259279720517194, 7.118632625653918)
t1, c, a, b = Polygon(A, B, C)
f, o_1 = AngularBisector(b, a)
h, o_2 = AngularBisector(c, a)
O = Intersect(o_1, o_2)
k = Circle(A, C, B)
ω = Circle(A, O, B)
S_1 = Center(ω)
g = OrthogonalLine(O, a)
T_a = Intersect(g, a)
e = Circle(O, T_a)
γ = Angle(A, C, S_1)
β_{1} = Angle(S_1, C, B)
D = Center(k)
p = CircleArc(D, A, S_1)
q = CircleArc(D, S_1, B)
j = Segment(A, S_1)
l = Segment(S_1, B)
β = Angle(O, B, C)
φ = Angle(S_1, O, B)
α = Angle(A, B, S_1)
i = Segment(S_1, B)
δ = Angle(O, B, A)
ε = Angle(O, B, S_1)
m = Segment(O, S_1)