Some further properties of this transformation are:

(1) The projections {Y,Z} of X and {Y',Z'} of its transform X' on the sides are on a circle c.

(2) The center of this circle c is on the bisector AD.

(3) The triangles XYZ and X'Y'Z' are equal.

(4) If a line AX is the locus of points such that XY/XZ = k, then its transform AX' is the locus of points X' such that X'Y'/X'Z' = 1/k.

Start from (3) showing the equality of triangles XYZ and X'Y'Z'. This is immediate, since by definition the angles at X, X' are equal and complementary to A and the sides XY=X'Z', XZ=X'Y'.

(4) is a consequence of (3). Properties (1) and (2) follow from the fact that YY'ZZ' is an equilateral trapezium, since AY'=AZ etc..

(1) Triangles XYZ and X'Y'Z' are similar.

(2) The quadrilateral YY'Z'Z is cyclic on a circle c, and {Y'Z, YZ'} are antiparallels.

(3) AE is orthogonal to Y'Z' and AE' is orthogonal to YZ.

(4) The center of the circle c is the middle O of XX'.

(5) The three lines {Y'Z, YZ', XX'} pass through the same point G, whose polar with respect to the sides {AB, AC} is the orthogonal from A to XX'.

(1) is valid because quadrangles AYXZ and AY'X'Z' are cyclic, thus angle YAX = YZX, X'Y'Z' = X'AZ' etc..(3) follows also from these angle equalities e.g. X'Y'Z' = YAX and since the sides {AY, X'Y'} are orthogonal the other two sides {XA,Y'Z'} are also orthogonal. (2) follows also from the equality of the angles YY'Z' = YY'X'+X'Y'Z' and YZZ'=YZX+XZZ' but X'Y'Z' = YXZ etc.. (4) follows from the fact that O projects on the middle of YY' on AB and the middle of ZZ' on AC, hence it is the intersection of the two medial lines of these segments of the same circle c.

(5) is proved in section-2 of Pedal.html .

(1) The result of applying the transformations successively to a point X, creating X'=F

X''=F

(2) The composition F=F

(1) is obvious since reflexions preserve distances, hence all IX, IX', IX'', etc. are equal.

(2) follows from (1). In fact, angle X'XX''' is then complementary to X'X''X''', which in turn has its legs orthogonal to the bisectors IB, IC. By measuring angle it follows that XX''' is parallel to AC. Since then XIX''' is isosceles X and X''' are symmetric with respect to the line from I which is orthogonal to AC.

playing an essential role in the geometry of the triangle and being quadratic in nature and transforming lines to conics passing through the vertices of the triangle (see Isogonal_Conjugation.html ).

Isogonal_Conjugation.html

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