Geom-examples

Tetsuo Ida and Hidekazu Takahashi
In[]:=
EosSession["Geometry Examples"];

Equilateral Triangle 1

Construction

In[]:=
NewOrigami[10,Context->"EquilateralTriangle1`"];
In[]:=
MarkOn[];
In[]:=
HO["A","D"]!
Geometry Examples/EquilateralTriangle1: Step 3
Out[]=
In[]:=
HO["D","EF","A",Mark{{"CD","W"}},FoldLine1]
Geometry Examples/EquilateralTriangle1: Step 4
Out[]=
In[]:=
Unfold[]
Geometry Examples/EquilateralTriangle1: Step 5
Out[]=
In[]:=
HO["C","EF","B",Mark{{"AW","G"}},FoldLine2]!
Geometry Examples/EquilateralTriangle1: Step 7
Out[]=
In[]:=
ShowOrigami[MarkPoints{"G","A","B","C","D"},More{Thickness[0.01],Hue[0],Line[{"A","B","G"}]},SphericalRegionFalse]
Geometry Examples/EquilateralTriangle1: Step 7
Out[]=

Verification

In[]:=
ProofDocFormat["Proof","Subsection",1];
In[]:=
ProofDocFormat["Goal","Subsubsection",1];
In[]:=
Goal[SquaredDistance["A","B"]==SquaredDistance["B","G"]==SquaredDistance["G","A"]];
In[]:=
Prove["Equilateral Triangle",ProofDocTrue]
Proof is successful.
Geometry Examples/EquilateralTriangle1: Step 7
Out[]=
{Success,0.009714,
Equilateral Triangle.pdoc.nb
}

Equilateral Triangle 2

Construction

In[]:=
NewOrigami[10,Context->"EquilateralTriangle2`"];
In[]:=
MarkOn[];
In[]:=
HO["A","B"]
Geometry Examples/EquilateralTriangle2: Step 2
Out[]=
In[]:=
HO["F","D"]
Geometry Examples/EquilateralTriangle2: Step 3
Out[]=
In[]:=
UnfoldAll[]//Column
Geometry Examples/EquilateralTriangle2: Step 5
Out[]=

Verification

Equilateral Triangle 3

This constriction is simplest. I found this recipe in Fushimi’s book (Geometry of Origami, 1979)

Construction

Verification

Equilateral Triangle 4

Construction

Verification

Orthocenter

The three (possibly extended) altitudes intersect in a single point H of the triangle.　Point H is called the orthocenter of the triangle.

Construction

Verification

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.

Construction

Verification

Circumcenter

For any triangle ΔABE, the perpendicular bisectors of the three edges AB, BE and EA meet at the same point H.
Point H is called the circumcenter of ΔABE.

Construction

Verification

Centroid 1

For any triangle ΔABE, the lines that pass through the midpoints of the three edges AB, BE and EA, and vertices E, A, and B, respectively, intersect at the same point G.
Point G is called the centroid (or the center of gravity) of ΔABE. The line that passes through the edge and the midpoint of the opposite side is called a median.

Construction

Suppose that we are given triangle ΔABE.

Verification

We prove the following
The three medians of any triangle all pass through one point. (see Coxeter P.10)
We can further prove that the medians trisects each other.

Centroid 2

Since the midpoint of the segment on the origami is foldable, Orikoto provides a function that gives the midpoint (P+Q)/2 of the segment PQ. Using the function Orikoto`Midpoint, we can construct the centroid as follows.

Construction

Verification

​

Excenter

For any triangle ΔABE, the angle bisectors of the three angles ∡E, π-∡A and π-∡B meet at the same point I.
Point I is called the excenter of ΔEAB.

Nine-point circle

The nine-point circle is a circle that can be constructed by the nine concyclic points defined from the triangle. These nine points are :
1
.
The midpoint of each side of the triangle.
2
.
The foot of each altitude.
3
.
The midpoint of the line segment from each vertex of the triangle to the orthocenter.

Construction

Point O is circumcenter

Verification

Ceva’s Theorem

Construction

Verification

Steiner-Lehmus Theorem

Construction

Verification

We prove that if AF = BG then AE =BE.

Menelaus’ Theorem

Construction

Verification

Euler line

Three centers are collinear:
For any triangle ΔABE, the circumcenter, the centroid and the orthocenter are collinear.
The line determined by the three points is called Euler line.

Construction

O is the circumcenter.

Simson Line Theorem

Given a triangle ABE and a point P on the circumcircle of the triangle, the three closest points on each line AB, BC and CE are collinear.

Construction

Verification

We prove that points X, Y and Z are collinear if P is on the circumcircle of ABE.

Load Eos