## 1. Similarly gliding

Given are three lines in general position {L1, L2, L3}. A triangle ABC glides with a vertex on each line,
while remaining similar to itself all the time. Then there is a point O, rigidly attached to the triangle,
remaining fixed all the time. Every point rigidly attached to the triangle glides also on a fixed line.

By the expression rigidly attached is meant a point which has the same (constant) trilinear coordinates  (see Trilinears.html ) with respect to
the varying ABC. Let {A0, B0, C0} be the intersections of lines {Li} and consider the three circles circumscribing respectively triangles
ABC0, BCA0, CAB0. The equality of angles (x) at A, B, C shows that the three circles pass through a point O. Projecting O on the lines
{Li} creates three similar right-angled triangles OAA', OBB', OCC'. The similarity of these three triangles implies the constancy
of O. In fact, relative to ABC, O is the point viewing the sides under the respective angles: BC under π-A0, CA under π-B0, AB under
π-C0. As such the point has ratios (OA:OB:OC) independent of the particular position of ABC. Later ratios equal (OA':OB':OC'), which
shows that O has fixed position with respect to A0B0C0. That every point rigidly attached to ABC glides then along a line follows from the,
basic for this subject, discussion in Similarly_Rotating.html .
Point O is called a pivot of rotation of  ABC in A0B0C0. There are 12 such nice points, which, like O, allow a similar rotation of ABC while
its vertices glide correspondinly on one of the three lines Li.

## 2. A case of duality

In sections 3 and 4 of Similarity.html , was examined a case, which in some sense is dual to the present one. A triangle ABC was there
restricted to move so that it always circumscribes a fixed triangle while remaining all the time similar to itself and each one of its vertices
is viewing a certain side of the fixed triangle under a constant angle. In this case it was shown that every point of the moving triangle,
which is rigidly attached to it, moves along a circle which passes through a fixed point O. This point was also called pivot of rotation (of the
moving outer triangle) because it coincides with the equal named point considered here. Simply instead to turn the outer triangle around, as
was the case there, here we turn the inscribed triangle around O, fixing the circumscribing triangle defined by the lines Li.
The difference of the two procedures is reflected in the way they vary all rigidly attached points to the moving triangle. In that case they
move along circles passing through O, in the present case they move along lines.

## 3. Special cases

From the previous discussion follows that given three lines {L1, L2, L3} in general position and a triangle ABC to be used as prototype.
There are twelve ways to let a similar to ABC triangle glide with its vertices lying all the time on these three lines. Isosceli though
reduce to six and equilaterals to two. In this later case the two pivots are the isodynamic points J, J' of the triangle ABC, having vertices the
intersections of the three lines. The figure below illustrates such a case (see Isodynamic.html and ApolloniusCircles.html ).

The above figure shows the two pivots creating all the various possible positions of an equilateral triangle whose vertices glide on the
three side-lines of triangle ABC.

Next figure illustrates another case, concerning the isosceles right-angled triangle. The six (red) points are all the pivots creating all possible
positions for the isosceles right-angled triangle to be inscribed in ABC. The pivots lie by three on two circles orthogonal to the bundle of
Apollonian circles of ABC. Each Apollonian circle contains two of them which are interchanged by the inversion with respect to the
circumcircle of ABC.

## 4. Application to inscribed squares

The last case of the isosceles right-angled triangle rotating inside another triangle can be used to construct all squares
inscribed in an arbitrary quadrilateral (i.e. a square having each vertex on a different side-line of a given quadrilateral).

For this it suffices to do the following:
[1] Single out three sides of the quadrilateral, {AB, BC, DA} say, making the triangle ABJ.
[2] Construct a pivot H of rotation of a right-angled isosceles EFG in ABJ.
[3] Complete EFG to a square EFGI and consider triangle HEI, having H fixed E on line AD and remaining
similar to itself. The locus of I, as HEI turns around H is a line (e).
[4] The intersection point K of (e) with the fourth side, CD, of the qudrilateral is a vertex of the inscribed square.
[5] To complete the square construction build on HK triangle HKL similar to HEI and construct on side KL the
square KLMN.
In general each of the six pivots of rotation of EFG in ABJ, playing the role of H, will deliver a different square and
we will have six solutions. Such a configuration of the six squares inscribed in a quadrilateral can be seen in
Inscribed_Squares_6.html .