### II. Case of n cocyclic points : Use of an inversion.

In this Paragraph, we suppose that points $A_i$ are on a circle $\Gamma$ of center O and radius *R* .*i* will stanf for the inversion of center O and ratio $R^2$.

Proposition 2.1 : We have T o i = i

**Demonstration :** Let’s have a point M of plane E and have N = *i* (M). The circle $\Gamma$ being invariant through *i*, proposition 1.4 gives us :

$N{A_i} = \frac{{R^2} M{A_i}}{OM \times O{A_i}} = \frac{R}{OM} \times MA_i$ .

The coefficients NA_{i} being proportional to coefficients MA_{i}, we deduct that the barycenter of points A_{i},MA_{i}) and (A_{i},NA_{i}) are identical, so we get *T*(M)=*T*(N)=(*T*o*i*)(M).

**Corollary 2.2 :** The image of plane E through *T* is equal to the image through *T* of the inside of the cercle of center O and radius *R* (circle included).

Indeed, if a point M is not inside the circle, it’s image through *i* is.

Continuation of the demonstration

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