## 1. Basic property for similar triangles

Given are a triangle t=(DEF), a fixed point C and a line e = AB. Triangle CGJ is moving so that it remains similar to
DEF, its vertex C is fixed, and vertex G moves on line e. Then its other vertex J is moving on a fixed line g.

The clue idea is to consider the circumcircle of the moving triangle. This intersects line AB at a fixed point K, with the
property angle(CKG)=angle(CJG) and angle(JKB)=angle(GCJ). Thus,  g may be constructed as follows: Let H be the
projection of C on e. On CH we construct triangle CHI, similar to DEF. g is the orthogonal to CI at I. An alternative
way would be to draw the triangle CGJ at the position which line GJ is identical with line AB. Then J obtains the
position of K and line g is the tangent at K of the circumcircle of CGK (figure below).

## 2. Extension to arbitrary shapes

The previous property can be extended to an arbitrary shape moving so that one of its points is fixed while the
shape remains similar to itself and some other point of it moves along a line AB. Then every other point of the shape
describes also a line.

The above figure shows such a case of a polygon EFGH which has E fixed, remains all the time similar to itself,
while a vertex (F) of it moves on a line. Then the other vertices {G, H} move on two lines correspondingly.

Remark The formulation of the property given above is more general than the property suggested by the
previous figure, which deals only with the vertices. The formulation relates to points rigidly attached to the shape
i.e. having fixed trilinear coordinates (see Trilinears.html ) with  respect to a triangle fixed in the shape and having all
the time the same form and position with respect to the shape. Thus, for example, the centroid of the shape describes
another fixed line.

## 3. Construction of triangle with vertices on three lines

To construct a triangle ABC similar to another A'B'C' and having its vertices correspondingly on three different lines {a,b,c}.

To solve the problem it suffices to take an arbitrary point B on b and carry out the previous construction, considering B
fixed and C moving on line c. Then A moves on a certain line a0, forming a fixed angle (π-B) at its intersection point K
with line c. Considering the intersection point of a0 with a we obtain a triangle ABC with the desired property. It is plain
that all triangles ABC having the vertices correspondingly A on a, B on b and C on c, can be obtained in this way.
In addition all points rigidly attached to the triangle (see previous remark) move also on corresponding lines.
In particular this happens with the circumcenter O of the triangle.
Regarding the behavior of the circumcircle the following cases can occur, when the three lines {a,b,c} pass through a
common point K:
[1] One of the circumcircles passes through this common point K. Then all circumcircles pass through this point
and we have a situation like that of the main proposition, where one of the vertices, B say, does not move really
on b, but remains fixed in place on it and only the other two points, A and C move on respective lines. In that
case lines a and c make an angle complementary to B.
[2] The second possibility results for triangles constructed as in [1] and letting also B vary on b through a homothety
transformation centered at K.
[3] The third possibility, when all three lines a, b, c pass through a point K, is when the circumcircle of ABC does
not pass through K (the angle at K is not (π-B)). In that case the angles of a0 with lines a,b,c remain fixed and
triangle ABC varies by homotheties centered at K. Thus, in this case the sides of all triangles ABC are parallel
to three respective directions.
In that case every point rigidly attached to ABC moves on a line passing through K.
Remark When the common point K of the three lines goes to infinity we obtain the limit case in which the
three lines a, b, c are parallel. In that case all triangles ABC resulting by continuously varying the vertices on the
respective lines are congruent.
A typical exercise along this lines would be:
Exercise Given three parallels a,b,c and a triangle A'B'C', construct a triangle ABC having its vertices on
these lines and being similar to A'B'C'.

Obviously we have six possibilities to select which point on which line to lie, plus the possibility to reverse the
orientation, which gives six additional cases resulting by reflecting the first to a line L orthogonal to the three parallels.

## 4. Relation to the pivot

The fixed point P about which is turning the shape while remaining all the time similar to itself is the so-called pivot
of the similarly rotating shape. The pivot can be discovered when we know the three lines {a,b,c} on which vary
three rigidly attached points {A,B,C} of the shape plus one single position for the (rigid) triangle of these points.
Then we have a configuration of a triangle ABC remaining similar to itself and having its three vertices vary correspondingly
on three lines {a,b,c}. This case is examined in the file SimilarlyGliding.html .

## 5. Similarities and magnification instruments

The property examined here is related to properties of (spiral) similarity transformations (see Similarity.html ) and can be
used to produce magnification instruments for drawing shapes in various scales.

Turning a triangle ABC about its vertex A so that it remains similar to itself means that AC/AB = k all the time and also
that angle(BAC) is constant. Thus the effect is the same with that of applying a similarity with center at A, angle
equal to angle(BAC) and ratio equal to k. If point B moves on the contour of a shape S, then C describes the image
contour of the transformed shape F(S), where F is the above similarity. The pivot point A corresponds to the
center of similarity.