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).

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.

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.

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 a

with line c. Considering the intersection point of a

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 a

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.

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:

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.

of the

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 .

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

SimilarlyGliding.html

Similarity.html

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