Proposition 1.1 : If MAB is a triangle et I the foot of the interior bisector form point M, the I belongs to segment [AB] and |

1.2

This rather basical property won’t be demonstrated here.

Proposition 1.2 (Ptolemy’ theorem) : If A, B, C et D are four points son aligned , then with equality if, and only of, the points are cocyclic and the quadrangle ABCD convex. |

This property won’t be demonstrated here.

Proposition 1.3 : If A et B are two distincts points and k a real number, k>0, the set containing the points M such as is :
The perpendicular bisector of segment [AB] if k = 1.
Else the circle of diameter [IJ] where I = bar (A,1); (B, k) et J = bar (A,1); (B, - k).More, if M is a some point of this circle non lying on line (AB), I et J are the repectives feet of the interior and exterior bisectors from M in triangle MAB. |

**Demonstration :**

If *k* = 1 :

M est sur la médiatrice [AB].

If :

.

So .

Naming I the barycenter of (A,1); (B, k) and J the barycenter of (A,1) et (B, - k) (these barycenters exist because and so et ) :

For each point M, and

So because and .

So M is on circle with diameter [IJ].

Considering a point M of this circle which does’nt belong to line (AB) and I’ the foot of the internal bisector of MAB from M. According to proposition 1.1, we know that . So point I’ is on the circle and on segment [AB]. Finally we get I’=I.

We deduct immediatly from this that J is the foot of the external bisector of M in triangle MAB since these two bisectors are perpendicular.

The figure below illustrates this property :

Proposition 1.4 : If i is thel’inversion of center O and ration k, then, for all points M and N with rrepective images M’ et N’ through i, we get : |

**Demonstration :**

We are in the complex plane with point O as origin.

If M, N, M’ et N’ have for affixes respectively , , , then :

and .

So

So we have

Proposition 1.5 : Let us consider a triangle ABC non isosceles in A, I and J the fett of the two bisectors of angle in A and K the midpoint of [IJ].Then line (AK) est tangent in A to the circumscribed to ABC. |

**Démonstration : **

We suppose that triangle ABC is not isosceles in A.

Let’s call the circumscribed circle of ABC, M the second intersection point of the internal bisector (AI) with the circle eand N the second intersection point of de la external bisecto (AJ) with circle .

The internal and externla bisectors being perpendicular, the triangle AMN is rectangle in A so it’s circumscribed circle (which is ) has for diameter segment [MN]. So segment [MN] is a diameter of .

Applying the inscribed angle theorem we deduct from it that and

(AI) is he internal bisector of angle A in so .

We deduct that

Having OM = OB = OC, triangles OMN et OMC are isometric so MB = MC. Knowing that OB = OC, (MO) is the perpendicualr bisector of [BC].

To demonstrate that (AO) is perpendicualr to (AK) we will demonstrate that .

K being the midpoint of [IJ], .

O being the midpoint of [MN], .

So we have .

Yet and .

Moreover car .

And .

We deduct so .

Proposition 1.6 : If T is the tangent in A to a circle of center O, then each circle with it’s center on T and going through A is unvariant by the inversion i of center O et ratio . |

**Demonstration : **We are in the complex plane with an orthonormal frame whose origin is O.

we will set , and *b* will stand for B affix.

The inversion *i* being involutive (*i* o *i*=*Id*), to demonstrate that it’s enough to show that .

A some point M of the circle of centre B and going through A has an affix as follows : awith real number.

It’s image through the inversion of center O and ration has an affix of this form : .

To show that M’ belongs to we just have to show that .

with .

.

So .

Yet because triangle OAB is rectangle in A (Pythagora theorem).

So .

Yet .

So and point M’ really belongs to the circle.