|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|
|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.
If k = 1 :
M est sur la médiatrice [AB].
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 :|
We are in the complex plane with point O as origin.
If M, N, M’ et N’ have for affixes respectively , , , then :
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.
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 .
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 .
Yet because triangle OAB is rectangle in A (Pythagora theorem).
So and point M’ really belongs to the circle.