A Similarity is a transformation F of the plane onto itself defined by a tripple (O, ω, k), consisting from a point Ï, an angle ω and a (positive) number k.Point O is the center of similarity and angle ω is the rotation angle of the similarity. Number k is the ratio of the similarity. The standard reference here is Yaglom's book [Yaglom, vol. II] which 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 P0 and then taking P' on OP0, such that OP'/OP0 = k.Thus, by definition, the similarity is a composition of two transformations: the rotation Rw and the homothety Hk.
It is easily shown that the order of factors in the composition is irrelevant.
A similarity transformation F is defined by giving two points {A,B} and prescribing their images {A'=F(A),B'=F(B)}. The center of similarity is the intersection point O of the circles through the tripples {A,A',I}, {B,B',I}, where I is the intersection point of the carrying lines of AB and A'B'. The ratio of similarity is the ratio of the lengths r = |A'B'|/|AB| and the rotation-angle of the similarity is the angle ω of the carrying 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 homothety with center at the intersection point of these lines. If segments {AA', BB'} are parallel and equal then O is at infinity and the transformation mapping AB to A'B' becomes a translation.
Remark A point X with given relative position with respect to AB (i.e. given k=XA/XB) maps by the 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 .
One of the most prominent examples in applying similarities, I think, is the following, in which a triangle circumscribes another remaining all the time similar to itself. To create the example take an arbitrary triangle t0 = DEF and draw on 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 fixed relative position of L is guaranteed by requiring that the projection OL of O on L is a point with fixed relative position with respect to the varying ABC. Then OL describes a circle through O and O* is the diametric of O on that circle. (The argument could be reversed, first proving this and then [5]).
O is called a pivot of rotation of a triangle similar to ABC about the triangle DEF. The trilinears of O (see 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.
Theorem If a triangle ABC changes position but remains similar to itself and its side-lines pass 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 OL, which is the diametric of O with respect to the circle described by the foot of the perpendigular to L from O.