angle ω and a (positive) number k.Point O is the

similarity. Number k is the

starts with an excellent review entitled "What is Geometry?".

For every point P different from O point F(P) = P' is defined by rotating P by the angle ω to point P

P' on OP

R

It is easily shown that the order of factors in the composition is irrelevant.

{A'=F(A),B'=F(B)}. The

{A,A',I}, {B,B',I}, where I is the intersection point of the carrying lines of AB and A'B'. The

the ratio of the lengths r = |A'B'|/|AB| and the

lines.

When the carrying lines are parallel point I goes to infinity, the circles become straight lines through {A,A'},

{B,B'} correspondingly, and the similarity becomes a

lines. If segments {AA', BB'} are parallel and equal then O is at infinity and the transformation mapping

AB to A'B' becomes a

similarity to point X' with the same relative position with respect to A'B' (i.e. X'A'/X'B' is also k). This has

the consequence that all circles {X,X',I} pass through O. There is also another point of view for this, discussed

in Thales_General.html .

another remaining all the time similar to itself. To create the example take an arbitrary triangle t

its sides circles the points of which (the outer arcs) view the sides under angles {φ, χ, ψ} with φ + χ + ψ = π.

Then take an arbitrary point A on the φ-arc and extend lines AE, AF until they cut again the corresponding circles at C, B.

The following facts follow directly from the definitions:

[1] Line BC passes from D and triangle t = ABC has angles {φ, χ, ψ}.

[2] The circles carrying A,B,C intersect at a common point O, characterized by the property to view the

sides of DEF under the complementary angles {π-φ, π-χ, π-ψ}.

[3] The center of similarity of two such triangles ABC and A'B'C' is point O.

[4] O has the same relative position with respect to all triangles ABC. This means that drawing the vertical

distances to the sides {OX, OY, OZ} the ratios between any pairs of them (OX : OY : OZ) are the same for all

triangles ABC. This because for two triangles ABC , A'B'C' the corresponding OX, OX' result by the same

similarity mapping ABC to A'B'C', hence their angle is the similarity angle and the ratio OX'/OX = r, where r is

the similarity ratio.

[5] Inversely, points U having the same relative position describe a circle passing through O. This can be

proved first for points U on the sides of ABC, by applying the above remark. Then it can be proved

for arbitrary points by considering the intersection points of OU with the sides and using again the

remark.

[6] The previous remark implies that lines L having fixed relative position with respect to ABC pass

through a point O*, the same for all these L (but changing if the relative position of L changes). In

fact, the

L is a point with fixed relative position with respect to the varying ABC. Then O

through O and O* is the diametric of O on that circle. (The argument could be reversed, first proving

this and then [5]).

Some further points on this subject are discussed in the files SimilarlyGliding.html , Similarly_Rotating.html ,

SimilarInscribed_0.html and SimilarInscribed.html and Pivot.html . The set of all similarities is not a group.

In order to become a group it has to be complemented with the set of translations. The reason is exposed in

Homotheties_Composition.html . Homotheties are similarities without rotation-part i.e. with w=0.

Trilinears.html ) relative to either of the two triangles (DEF and ABC) are uniquely determined by the angles of

the two triangles and the choice of which is opposite to which. Since we can pair any one of the angles {d,e,f}

of DEF with some of the angles {φ, χ, ψ} we come at six possible pivots. By allowing the vertices of

ABC to view the sides of DEF also under the complementary angles correspondingly {π-φ,π-χ,π-ψ} (see

image below) we come at twelve pivots relating two triangles with given angles. But this is another story (see

SixPivots.html ) handled usually from the (dual in some sense) point of view of turning DEF around its pivot in

ABC, so that it remains similar to itself all the time.

After this short digression on pivots, I come to my subject to state the results in the form of a theorem.

trhough three fixed points {D,E,F} then:

i) All similarities interchanging two such positions of ABC have the same center O.

ii) All points fixed relative to ABC describe circles through O.

iii) All lines L fixed relative to ABC pass through some point O

respect to the circle described by the foot of the perpendigular to L from O.

SimilarlyGliding.html

Similarly_Rotating.html

SimilarInscribed_0.html

SimilarInscribed.html

Pivot.html

Homotheties_Composition.html

Trilinears.html

SixPivots.html

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