157. Steinbart's theorem. Applications.

by Virgil Nicula, Oct 16, 2010, 7:40 AM

http://geoma29.wordpress.com/

Lemma 1. Let $ABCDEF$ be a cyclical hexagon (convex or not).

Then $\boxed{AD\cap BE\cap CF\ne\emptyset\ \iff\ AB\cdot CD\cdot EF=BC\cdot DE\cdot FA}$ .


Proof 1. Suppose that $I\in AD\cap BE\cap CF$ . From the product of evident relations $\frac {IA}{IE}=\frac {AB}{DE}\ \wedge\ \frac {IC}{IA}=\frac {CD}{FA}\ \wedge\ \frac {IE}{IC}=\frac {EF}{BC}$ obtain the required relation

$AB\cdot CD\cdot EF=BC\cdot DE\cdot FA\ (*)$ . Reciprocally, suppose that the relation $(*)$ is true. Denote $X\in AD\cap BE$ and $C_1\in XF\cap w$ ,

where $w$ is the circumcircle of the given hexagon. Appling the direct implication which was proved above obtain $AB\cdot C_1D\cdot EF=BC_1\cdot DE\cdot FA\ \ (1)$ .

From the relations $(*)$ and $(1)$ results $\frac {C_1D}{C_1B}=\frac {CD}{CB}$ . Since $\widehat{BC_1D}\equiv\widehat{BCD}$ obtain $\triangle BC_1D\sim\triangle BCD$ . Since these triangles have same circumcircle $w$

(similarity ratio is equal to $1$) obtain $\triangle BC_1D\equiv\triangle BCD$ , i.e. $BC_1=BC$ and $C_1D=CD$ . In conclusion, $C_1\equiv C$ , i.e. $AD\cap BE\cap CF\ne\emptyset$ .

Proof 2. Suppose w.l.o.g. that the quadrilaterals $ADEF$ and $ABCD$ are convex. Denote $X\in AD\cap BE$ , $Y\in AD\cap FC$ . Using the well-known relations

$\frac {XA}{XD}=\frac {AB\cdot AE}{DB\cdot DE}$ (in $ABDE$) and $\frac {YA}{YD}=\frac {AC\cdot AF}{DC\cdot DF}$ (in $ACDF$) obtain that $AD\cap BE\cap CF\ne\emptyset\iff$ $X\equiv Y\iff$ $\frac {XA}{XD}=\frac {YA}{YD}\iff$

$AB\cdot CD\cdot (AE\cdot DF)=$ $DE\cdot AF\cdot (AC\cdot BD)$ . Using the Ptolemy's relation in the cyclical quadrilaterals $ADEF$ and $ABCD$ obtain the relations

$AE\cdot DF=AD\cdot EF+AF\cdot DE$ and $AC\cdot BD=AB\cdot CD+AD\cdot BC$ . Therefore, $X\equiv Y\iff$ $AB\cdot CD\cdot (AD\cdot EF+AF\cdot DE)=$

$DE\cdot AF\cdot (AB\cdot CD+AD\cdot BC)\iff$ $AB\cdot CD\cdot AD\cdot EF=$ $DE\cdot AF\cdot AD\cdot BC\iff$ $AB\cdot CD\cdot EF=BC\cdot DE\cdot FA$ .

Remark. Using the previous result for the cyclical quadrilaterals $ACBDFE$ and $BACEDF$ obtain the following chain of equivalencies :


$\left\|\begin{array}{cccc}
ABCDEF\ : & AD\cap BE\cap CF\ne\emptyset & \iff & AB\cdot  CD\cdot EF=BC\cdot DE\cdot FA\\\\
ACBDFE\ : & AD\cap CF\cap BE\ne\emptyset & \iff & AC\cdot BD\cdot FE=CB\cdot  DF\cdot EA\\\\
BACEDF\ : & BE\cap AD\cap CF\ne\emptyset & \iff & BA\cdot CE\cdot  DF=AC\cdot ED\cdot FB\end{array}\right\|$ .


Lemma 2. Let $AE$ , $AF$ be the tangent lines from $A$ to the circle $w$ , where $\{E,F\}\subset w$ . Consider the points $D$ , $D'$ so $EF$

separates $A$ , $D$ and doesn't separate $A$ , $D'$ . Denote $\left\|\begin{array}{c}
m(\widehat {D'AF})=x\\\
m(\widehat {D'AE})=y\end{array}\right\|$ . Then exists the relation $\frac {\sin x}{\sin y}=\left(\frac {D'F}{D'E}\right)^2$ .


Proof. Denote $\left\|\begin{array}{c}
m(\widehat {D'FA})=u\\\
m(\widehat {D'EA})=v\end{array}\right\|$ . Apply the Sinus' theorem in the triangles $AED'$ and $AFD'$ , i.e. $\left\|\begin{array}{c}
\frac {FD'}{\sin x}=\frac {AD'}{\sin u}\\\\
\frac {AD'}{\sin v}=\frac {ED'}{\sin y}\\\\
\frac {\sin v}{\sin u}=\frac {ED'}{FD'}\end{array}\right\|\ \bigodot\ \implies\ \frac {\sin x}{\sin y}=\left(\frac {D'F}{D'E}\right)^2$ .

Lemma 3. Let $D'$ , $E'$ , $F'$ be three points which belong to the outside of the triangle $ABC$ . Denote $\left\|\begin{array}{c}
m(\widehat{D'AB})=x_1\ ,\ m(\widehat{D'AC})=x_2\\\
m(\widehat{E'BC})=y_1\ ,\ m(\widehat{E'BA})=y_2\\\
m(\widehat{F'CA})=z_1\ ,\ m(\widehat{F'CB})=z_2\end{array}\right\|$ .

Then $\boxed{AD'\cap BE'\cap CF'\ne\emptyset\ \iff\ \frac {\sin x_1}{\sin x_2}\cdot\frac {\sin y_1}{sin y_2}\cdot\frac {\sin z_1}{\sin z_2}=1}$
(it is the trigonometric form of the Ceva's theorem).



Steinbart's theorem. Let $ABC$ be a triangle with the incircle $w$ which touches it at $D\in BC$ , $E\in CA$ , $F\in AB$ .

Consider three points $\{D',E',F'\}\subset w$ . Then $\boxed{DD'\cap EE'\cap FF'\ne\emptyset\ \iff\ AD'\cap BE'\cap CF'\ne\emptyset}$ .



Proof. Using the lemma 2 obtain $\left\|\begin{array}{c}
\frac {\sin x_1}{\sin x_2}=\left(\frac {D'F}{D'E}\right)^2\\\\
\frac {\sin y_1}{\sin y_2}=\left(\frac {E'D}{E'F}\right)^2\\\\
\frac {\sin z_1}{\sin z_2}=\left(\frac {F'E}{F'D}\right)^2\end{array}\right\|\ \bigodot\ \implies$ $\frac {\sin x_1}{\sin x_2}$ $\cdot\frac {\sin y_1}{\sin y_2}\cdot\frac {\sin z_1}{\sin z_2}=$ $\left(\frac {D'F}{D'E}\cdot\frac {E'D}{E'F}\cdot\frac {F'E}{F'D}\right)^2\ \ (1)$ .

Apply the lemma 2 to the cyclical hexagon $D'FE'DF'E$ , i.e. $D'D\cap FF'\cap E'E\ne\emptyset\ \implies\ D'F\cdot E'D\cdot F'E=FE'\cdot DF'\cdot ED'$ .

Thus the relation $(1)$ becomes $\frac {\sin x_1}{\sin x_2}$ $\cdot\frac {\sin y_1}{\sin y_2}\cdot\frac {\sin z_1}{\sin z_2}=1$ . Using the last lemma obtain that $AD'\cap BE'\cap CF'\ne\emptyset$ .



Applications.

PP1.The incircle of $\triangle ABC$ touches its side $BC$ at $K$. Let $M$ be the midpoint of the altitude $AD$ of triangle $ABC\ ,\ D\in BC$ . The line $MK$ meets

the incircle $w$ of $\triangle ABC$ at a point $N$ (apart from $K$). Show that the circumcircle of $\triangle BNC$ is tangent to the incircle of $\triangle ABC$ at the point $N$.


Proof. Suppose w.l.o.g. $c<b$ . Denote the A-exincircle $w_a=C(I_{a},r_{a})$ , the point $D'\in BC\cap w$, the intersection $P$ between the line $BC$ with the tangent

to the incircle in the point $N$ and $R\in NK\cap IP$ . Prove easily that $KD'=b-c$, $KD=\frac{(b-c)(p-a)}{a}$ and $\frac{KD}{KD'}=\frac{p-a}{a}$. Since $\frac{DM}{D'I_{a}}=$ $\frac{h_{a}}{2r_{a}}=$

$\frac{ah_{a}}{2ar_{a}}=$ $\frac{S}{ar_{a}}=$ $\frac{r_{a}(p-a)}{ar_{a}}=$ $\frac{p-a}{a}$ obtain that $\frac{DM}{D'I_{a}}=\frac{KD}{KD'}$, i.e. $\boxed{\ I_{a}\in KM\ }$ . The points $B,C,R$ belong to the circle with the diameter $II_{a}$. Since

$IP\perp NK$ obtain that $PN^{2}=PR\cdot PI=$ $PB\cdot PC$. Thus $PN^{2}=PB\cdot PC$, i.e. the line $PN$ is tangent to the circumcircle of the triangle $BNC$ .


PP2. Let $\gamma$ be the incircle in the triangle $A_0A_1A_2$. For all $i\in\{0,1,2\}$ we make the following constructions (all indices are considered modulo 3):

$\gamma_i$ is the circle tangent to $\gamma$ which passes through the points $A_{i+1}$ and $A_{i+2}$; $T_i$ is the point of tangency between $\gamma_i$ and $\gamma$; finally, the common

tangent in $T_i$ of $\gamma_i$ and $\gamma$ intersects the line $A_{i+1}A_{i+2}$ in the point $P_i$. Prove that $P_0\in P_1P_2$ and $A_0T_0\cap A_1T_1\cap A_2T_2\ne\emptyset$ .


Proof.
This post has been edited 50 times. Last edited by Virgil Nicula, Dec 1, 2015, 10:12 AM

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