In this section, we consider deux points A and B distincts of plane E and we study l’The transformation *T* defined by *T*(M)= bar (A,MA); (B,MB).

**1. Image of T.**

MA et MB being positives, *T* takes it’s values in segment [AB]. *T* is obviously surjective because, in restriction to segment [AB], *T* is the symmetry through I, midpoint of [AB].

**2. Geometrical construction of the image through T of a point M non aligned with A and B.**

We consider $\Gamma$ the circumscribed circle of triangle MAB. Let’s name P the intersection point of the perpendicular bisector of segment [AB] with the circle arc of summits A and B which does’nt contain M, and N the intersection point of line (MP) with segment [AB]. Then we know that N is the foot of the internal bisector of M in the triangle AMB. Proposition 1.1 implies that $\frac{MA}{MB} = \frac{NA}{NB}$ . N being the barycentre of (A, NB) and (B, NA), we deduct that N is the barycenter of (A, MB) and (B, MA). The symmetric point M ’ of M through I (midpoint of [AB]) is therefore the barycenter of (A, MA)and (B, MB). In other words : T (M) = M’.The inverse construction prooves that if $\Gamma$ is a circle going through A and B and M ’ a point of segment ]AB[, There is an unique point M of the circle arc of $\Gamma$ which does’nt contain P such as M ’= T(M). |

In other words :

**Proposition 3.1 :** If $\Gamma$ is a circle and A and B two distincts points of this circle, the transformation *T* sets une bijection of each of the two circle arcs of $\Gamma$ with summits A et B in the segment [AB].

**3. Fibers of T.**

We will call fibers of *T* the sets $T^{ - 1} \left( {\left\{ C \right\}} \right)$ for C $\in$ [AB].

When C = A (resp. C = B) ,The fiber of C is reduced in {B} (resp. to {A}).

If C $\in$ [AB], C $\ne$ A and C $\ne$ B, then C is barycenter of (A, $\lambda$) and (B, $1-\lambda$) with 0 < $\lambda$ < 1.

If M $\in$ E - {A, B}, *T* (M) = C is equivalent to say that the barycenter of points (A, MA) et (B, MB) and the barycenter of (A, $\lambda$) and (B, $1-\lambda$) are the same, which is true if, and only if, $\frac{MA}{MB} = \frac{\lambda }{{1 - \lambda }}$ .

So the fiber of C is the set $C_{A,B,\frac{\lambda }{1 - \lambda }}$.

We have therefore a criterion for *T* to be injective when applied to a set $\Gamma$ : the ecessary and sufficient condition is that $\Gamma$ don’t intercept two times the circles C_{A,B,k} with *k*>0. It is the case when we have an arc of circle joining A and B, but also in many other occasions : an arc of ellipsa, a broken line [AC] $\cup$ [CB] where C is a point of the perpendicular bisector of [AB], ot, as shown on the figure above, a well chosen Bezier curve .

Continuation of the demonstration

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