Incenter
Incenter
Load Eos
Load Eos
In[]:=
<<EosLoader`
Eos3.7.1.1 (March 21,2023) running under Mathematica 13.2.0 for Mac OS X ARM (64-bit) (November 18, 2022) on Thu 30 Mar 2023 14:40:55.
Incenter
Incenter
For any triangle ΔABE, the angle bisectors of the three angles ∡A, ∡B and ∡E meet at the same point I.
Point I is the incenter of ΔABE.
In[]:=
EosSession["Incenter"];
In[]:=
NewOrigami[10]
Incenter: Step 1
Out[]=
In[]:=
NewPoint["E"{9,8}]
Incenter: Step 1
Out[]=
In[]:=
HO["AE"]!
Incenter: Step 3
Out[]=
In[]:=
HO["AE","AB"]
Incenter: Step 3
Out[]=
,
In[]:=
HO["AE","AB",FoldLine1]!
Incenter: Step 5
Out[]=
In[]:=
HO["BE"]!
Incenter: Step 7
Out[]=
In[]:=
HO["BA","BE"]
Incenter: Step 7
Out[]=
,
In[]:=
HO["BA","BE",FoldLine2,Mark{{$fline[4],"I"}}]!
Incenter: Step 9
Out[]=
In[]:=
HO["AB","I",Mark{{"AB","F"}}]!;
In[]:=
triangle={Thick,Red,Line[{"A","B","E","A"}]};
In[]:=
circle={Thick,Green,GraphicsCircle["I","IF"]};
In[]:=
ShowOrigami[More{triangle,circle}]
Incenter: Step 11
Out[]=
In[]:=
(*proofworksfromtheversionEos3.3.1*)
In[]:=
Goal[O3Q["EA","EB","EI"]SquaredDistance["I","F"]==SquaredDistanceLine["I","EB"]==SquaredDistanceLine["I","EA"]SquaredDistanceLine["I","AB"]];
In[]:=
Prove["Incenter",Mapping{"A"{0,0},"B"{1,0},(*"C"{1,1},"D"{0,1},*)"E"{u,v}}]
Proof is successful.
Incenter: Step 11
Out[]=
In[]:=
EndSession[];