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New theorem based on Ptolemy’s theorem : Preliminary results

publication Thursday 14 August 2014.

In this article we will see a few results we will later use for the new theorem

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 $\frac{MA}{MB} = \frac{IA}{IB}$


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 $AC \times BD \le AB \times CD + BC \times DA$ 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 $C_{A,B,k}$ containing the points M such as $\frac{MA}{MB}=k$ 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 :
$\frac{MA}{MB} = k \Leftrightarrow \frac{MA}{MB} = 1 \Leftrightarrow MA = MB \Leftrightarrow$ M est sur la médiatrice [AB].
- If $k \ne 1$ :
$\frac{MA}{MB} = k \Leftrightarrow MA = kMB \Leftrightarrow M{A^2} = {\left( {kMB} \right)^2} \Leftrightarrow {\left( {\overrightarrow {MA} } \right)^2} = {\left( {k\overrightarrow {MB} } \right)^2}$.
So $ \frac{MA}{MB} = k \Leftrightarrow {\left( {\overrightarrow {MA} } \right)^2} - {\left( {k\overrightarrow {MB} } \right)^2} = 0 \Leftrightarrow \left( {\overrightarrow {MA} + k\overrightarrow {MB} } \right).\left( {\overrightarrow {MA} - k\overrightarrow {MB} } \right) = 0$.

Naming I the barycenter of (A,1); (B, k) and J the barycenter of (A,1) et (B, - k) (these barycenters exist because $ k > 0 $ and $ k \ne 1$ so $ 1+k \ne 0$ et $ 1-k \ne 0$ ) :
For each point M, $\overrightarrow {MA} + k\overrightarrow {MB} = \left( {1 + k} \right)\overrightarrow {MI} $ and $\overrightarrow {MA} - k\overrightarrow {MB} = \left( {1 - k} \right)\overrightarrow {MJ} $
So $\frac{MA}{MB} = k \Leftrightarrow (1 + k)\overrightarrow {MI} .(1 - k)\overrightarrow {MJ} = 0 \Leftrightarrow \overrightarrow {MI} .\overrightarrow {MJ} = 0$ because $ 1+k \ne 0$ and $ 1-k \ne 0$.
So $\frac{{MA}}{{MB}} = k \Leftrightarrow $ 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 $\frac{I’A}{I’B}=\frac{MA}{MB}=k$. 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 : $M’N’ = \frac{{\left| k \right|MN}}{{OM\;ON}}$

Demonstration :
We are in the complex plane with point O as origin.
If M, N, M’ et N’ have for affixes respectively $\alpha$, $\beta$, $\alpha’$, $\beta’$ then :
$\alpha ’ = \frac{k}{\overline \alpha }$ and $\beta ’ = \frac{k}{\overline \beta }$.

So $M’N’=\left| \beta’-\alpha’ \right|=\left| \frac{k}{\overline{\beta}}-\frac{k}{\overline {\alpha}} \right|=\left| k \right|\times \left| \frac{1}{\overline {\beta}} - \frac{1}{\overline {\alpha }}\right|=\left| k \right|\times \left| \frac{\overline{\alpha}-\overline{\beta}}{\overline{\alpha}\overline{\beta}} \right|=\left| k \right|\times \frac{\left| \overline{\alpha-\beta} \right|}{\left| \alpha \beta \right|}=\left| k \right|\times \frac{\left| \alpha -\beta \right|}{\left| \alpha \beta \right|}$

So we have $M’N’=\frac{\left| k \right|\times \left| \alpha -\beta \right|}{\left| \alpha \right|\times \left| \beta \right|}=\frac{\left| k \right|\times MN}{OM\times ON}$

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$\Gamma$ the circumscribed circle of ABC, M the second intersection point of the internal bisector (AI) with the circle $\Gamma$ eand N the second intersection point of de la external bisecto (AJ) with circle $\Gamma$.

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

Applying the inscribed angle theorem we deduct from it that $\widehat {BMN} = \widehat {BAN} = \widehat {BAI}$ and $\widehat {CMN} = \widehat {CAN} = \widehat {CAI}$

(AI) is he internal bisector of angle A in so $\widehat {BMN} = \widehat {BAI} = \widehat {CAI}$.
We deduct that $\widehat {BMN} = \widehat {CMN}$

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 $\overrightarrow {AO} .\overrightarrow {AK} = 0$.

K being the midpoint of [IJ], $\overrightarrow {AK} = \frac{1}{2}\left( {\overrightarrow {AI} + \overrightarrow {AJ} } \right)$.

O being the midpoint of [MN], $\overrightarrow {AO} = \frac{1}{2}\left( {\overrightarrow {AM} + \overrightarrow {AN} } \right)$.
So we have $\overrightarrow {AO} .\overrightarrow {AK} = \frac{1}{4}\left( {\overrightarrow {AM} + \overrightarrow {AN} } \right).\left( {\overrightarrow {AI} + \overrightarrow {AJ} } \right) = \frac{1}{4}\left( {\overrightarrow {AM} .\overrightarrow {AI} + \overrightarrow {AN} .\overrightarrow {AI} + \overrightarrow {AM} .\overrightarrow {AJ} + \overrightarrow {AN} .\overrightarrow {AJ} } \right)$.
Yet $\overrightarrow {AM} .\overrightarrow {AI} = 0$and $\overrightarrow {AN} .\overrightarrow {AJ} = 0$.
Moreover $\overrightarrow {AN} .\overrightarrow {AI} = \left( {\overrightarrow {AM} + \overrightarrow {MN} } \right).\overrightarrow {AI} = \overrightarrow {AM} .\overrightarrow {AI} + \overrightarrow {MN.} \overrightarrow {AI} = \overrightarrow {MN.} \overrightarrow {AI} $ car $(AM) \bot \left( {AI} \right)$.
And $\overrightarrow {AM} .\overrightarrow {AJ} = \left( {\overrightarrow {AN} + \overrightarrow {NM} } \right).\overrightarrow {AJ} = \overrightarrow {AN} .\overrightarrow {AJ} + \overrightarrow {NM} .\overrightarrow {AJ} = 0 + \overrightarrow {NM} .\overrightarrow {AJ} = \overrightarrow {NM} .\left( {\overrightarrow {AI} + \overrightarrow {IJ} } \right) = \overrightarrow {NM} .\overrightarrow {AI} + \overrightarrow {NM} .\overrightarrow {IJ} = \overrightarrow {NM} .\overrightarrow {AI} $.
We deduct $\overrightarrow {AO} .\overrightarrow {AK} = \frac{1}{4}\left( {\overrightarrow {MN} .\overrightarrow {AI} + \overrightarrow {NM} .\overrightarrow {AI} } \right) = 0$ so $(AO) \bot (AK)$.

Proposition 1.6 : If T is the tangent in A to a circle $\Gamma$ 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 $OA^2$.

Demonstration : We are in the complex plane with an orthonormal frame whose origin is O.
we will set $OA=R$, $AB=r$ and b will stand for B affix.
The inversion i being involutive (i o i=Id), to demonstrate that $i\left( \Gamma \right) = \Gamma $ it’s enough to show that $i\left( \Gamma \right) \subset \Gamma $.
A some point M of the circle of centre B and going through A has an affix as follows : $z = b + re^{i\theta }$ awith $\theta$ real number.
It’s image through the inversion of center O and ration $R^2$ has an affix of this form : $z’=\frac{R^2}{\overline {z}}$.

To show that M’ belongs to $\Gamma$ we just have to show that $\left| z’ - b \right| = r$.

$\left| z’-b \right|=\left| \frac{R^2}{\overline {z}}-b \right|=\left| \frac{R^2-b\overline {z}}{\overline {z}} \right|$ with $\overline {z}=\overline {b+r\mathrm{e}^{i\theta }}=\overline {b}+r\mathrm{e}^{-i\theta }$.

So $z’-b=\frac{R^2-b\times \left( \overline {b}+r\mathrm{e}^{-i\theta } \right)}{\overline {b}+r\mathrm{e}^{-i\theta }}=\frac{R^2-b\overline {b}-br\mathrm{e}^{-i\theta}}{\overline {b}+r\mathrm{e}^{i\theta}}=\frac{R^2-{\left| b \right|}^2-br\mathrm{e}^{-i\theta}}{\overline {b}+r\mathrm{e}^{i\theta}}$.

Yet ${\left| b \right|^2} = O{B^2} = O{A^2} + A{B^2} = {R^2} + {r^2}$ because triangle OAB is rectangle in A (Pythagora theorem).

So $\left| z’-b \right|=\left| \frac{R^2-\left( R^2+r^2 \right)-br\mathrm{e}^{-i\theta }}{\overline {b}+r\mathrm{e}^{-i\theta}} \right|=\left| \frac{-r\left( r+b\mathrm{e}^{-i\theta} \right)}{\overline {b}+r\mathrm{e}^{-i\theta}} \right|=\left| \frac{-r\mathrm{e}^{i\theta}\left( r\mathrm{e}^{i\theta}+b \right)}{\overline {b}+r\mathrm{e}^{-i\theta}} \right|=r\left| \frac{r\mathrm{e}^{-i\theta}+b}{\overline {b}+r\mathrm{e}^{-i\theta}} \right|$.

Yet $\left| \frac{r\mathrm{e}^{i\theta}+b}{\overline b+r\mathrm{e}^{-i\theta }} \right|=\left| \frac{r\mathrm{e}^{i\theta}+b}{\overline{r\mathrm{e}^{i\theta}+b}} \right|=\frac{\left| r\mathrm{e}^{i\theta }+b \right|}{\left| \overline{r\mathrm{e}^{i\theta}+b} \right|}=\frac{\left| r\mathrm{e}^{i\theta}+b \right|}{\left| r\mathrm{e}^{i\theta}+b \right|}=1$.

So $\left| z’-b \right| = r$ and point M’ really belongs to the circle$\Gamma$.

Continuation of the demonstration

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