Consider a triangle ABC and the reflexion of points on a bisector AD of angle A say. This transformation
is called isogonal with respect to the bisector AD. The transformation, being a particular reflexion
preserves distances and maps lines to lines. In particular a line AX through A is mapped to a line AX'
through A, which is fixed by the transformation. Also sides AB and AC are interchanged by the transformation. 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..
Consider two isogonal lines AE, AE' of the triangle ABC and take an arbitrary point X on AE and
an arbitrary point X' on AE'. Then project X, X' on the sides of the triangle to points {Y,Y'} on AB
and {Z,Z'} on AC [Lalesco, p.40]. (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 . Remark In the aforementioned reference it is also proved that the polar of G with respect to
the pair of lines AB, AC passes through the intersection point of lines {YZ, Y'Z'}.
Consider a triangle ABC and the three reflexions FA, FB, FC with respect to its bisectors at the
corresponding angles. (1) The result of applying the transformations successively to a point X, creating X'=FA(X), X''=FB(X'), X'''=FC(X''), defines a cyclic quadrangle XX'X''X''' centered at I, the incenter of
the triangle. (2) The composition F=FC*FB*FA is the reflexion on the line through I which is orthogonal to AC.
(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. Remark-1 This property of the composition of reflexions is a special case of composition of
several reflexions on lines passing through a point. If the number of different lines is even then the
result-composition is a rotation if the number of lines is odd (as is here the case), then the
result is a reflexion. Remark-2 This composition of isogonal conjugation must be distinguished from the
isogonal conjugation with respect to a triangle which is another kind of important transformation, 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 ).