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A not difficult concurrency.

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#1 h Jun 20, 2007, 3:48 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

A triangle $\bigtriangleup ABC$ is given, and let the external angle bisector of the angle $\angle A$ intersect the lines perpendicular to $BC$ and passing through $B$ and $C$ at the points $D$ and $E$ , respectively. Prove that the line segments $BE$ , $CD$ , $AO$ are concurrent, where $O$ is the circumcenter of $\bigtriangleup ABC$ .

Kostas Vittas.

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#2 h Jun 20, 2007, 4:28 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

in triangle $\triangle BAC$ the lines $BE,CE,AE$ are concurrent at point $E$ so we get that:

$\frac{\sin ABE}{\sin EBC}.\frac{\sin BCE}{\sin ECA}.\frac{\sin EAC}{\sin EAB}=1$

so by putting $\angle ABE=\angle B_{1}$ and $\angle EBC=\angle B_{2}$ we get that:

$\frac{\sin B_{1}}{\sin B_{2}}=\frac{\cos C.\cos{\frac{A}{2}}}{\cos{\frac{A}{2}}}$

so:

$\frac{\sin B_{1}}{\sin B_{2}}=\cos C$ $(I)$

also same wise we get that:

$\frac{\sin C_{1}}{\sin C_{2}}=\cos B$ $(II)$

where $\angle C_{1}=\angle DCA$ and $\angle C_{2}=\angle DCB$

now for proving that the three desired lines are concurrent we have to prove that:

$\frac{\sin BAO}{\sin OAC}.\frac{\sin ACD}{\sin DCB}.\frac{\sin EBC}{\sin EBA}=1$

or we have to prove that:

$\frac{\cos C}{\cos B}.\frac{\sin C_{1}}{\sin C_{2}}.\frac{\sin B_{2}}{\sin B_{1}}=1$

which is trivialy true by $(I),(II)$

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#3 h Jun 20, 2007, 6:35 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Nice problem, vittasko! A synthetic proof should be the following:

Denote $K \in AA \cap BC$ , $G, F$ the points where the internal, respective external bisector of $\angle{BAC}$ cuts $BC$ , therefore $KA^{2}=KB \cdot KC =\frac{a^{2}b^{2}c^{2}}{(b^{2}-c^{2})^{2}}= (\frac{ac^{2}}{b^{2}-c^{2}}+\frac{ac}{b+c})^{2}= (KB+GB)^{2}= KG^{2}$ , so $KA=KG$ .

So $\angle{KAG}=\angle{KGA}$ $(1)$ .

But $\angle{KGA}=\angle{FGA}=\angle{FEC}=90-\angle{CFE}$ . $(2)$ .

Now by Menelaus applyed in $\triangle{FBX}$ for the transversal $C-H-E$ , where $H \in FX \cap CE$ we obtain:

$\frac{FC}{CB}\cdot \frac{BE}{EX}\cdot \frac{XH}{HF}=1$ .

But $\frac{XH}{HF}=\frac{CG}{CF}$ , because $XG \perp BC$ , so $\frac{BE}{EX}=\frac{BC}{CG}$ .

On other hand, $\frac{BC}{CG}=\frac{BF}{FK}\iff BC(FB-BK)=FB(BC-BG)$

$\iff BC \cdot BK=BF \cdot BG \iff a \cdot \frac{ac^{2}}{b^{2}-c^{2}}= \frac{ac}{b-c}\cdot \frac{ac}{b+c}$ , which is obvious.

So $\frac{BE}{EX}=\frac{BF}{FK}$ , thus $KX \| EF$ , so $\angle{CFE}=\angle{XCG}$ .

And by replacing in $(2)$ we obtain that $\angle{KGA}=90-\angle{XCG}=\angle{KXG}$ .

Therefore by replacing now in $(1)$ , $\angle{KAG}=\angle{KXG}$ , so the quadrilateral $KAXG$ is cyclic, thus $AX \perp AK$ , but $AO \perp AK$ , therefore $A,X,O$ are collinear, i.e the lines $BE, CD$ and $AO$ are concurrent.

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#4 h Jun 20, 2007, 8:58 PM • 5 Y

Y by Adventure10, Mango247, and 3 other users

This is a particular case of the Jacobi theorem, because:

Let X be the point diametrically opposite to A on the circumcircle of triangle ABC. Then, AX is a diameter of this circumcircle, so that < ABX = 90°. Thus, using < DBC = 90°, we get < CBX = < ABX - < ABC = 90° - < ABC = < DBC - < ABC = < ABD. Similarly, < BCX = < ACE.

Now, we have:

< EAC = < BAD (since the line DE is the exterior angle bisector of the angle A);
< DBA = < CBX (this is just a restatement of < CBX = < DBA);
< XCB = < ACE (this is a restatement of < BCX = < ACE).

Hence, the Jacobi theorem yields that the lines AX, BE and CD concur. Now, the line AX coincides with the line AO (since the segment AX is a diameter of the circumcircle of triangle ABC, and thus it passes through the center O of this circumcircle). Hence, the lines AO, BE and CD concur, qed.

PS. I have slightly edited your post in the hope of making it better to understand.

PPS. Here is another neat application of the Jacobi theorem: http://www.mathlinks.ro/Forum/viewtopic.php?t=130813 .

Darij

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#5 h Jun 20, 2007, 10:07 PM • 2 Y

Y by Adventure10, Mango247

Thank you dear friends BaBak Ghallebi, Pohoatza and Darij, for your interest and solutions.

I like better the synthetic solutions and the Darij's one, is the simplest ( of two ) I have in mind as a direct application of the Jacobi theorem $($ I like better to say the isogonic theorem, because it is also true in more its general configuration, with respect to the isogonal lines through vertices of a given triangle and I think that it is a good time to post here as soon is posible, the two elementary proofs I have in mind $).$

The Pohoatza's solution, is new to me, and I will check it carefully before a definite answer ( I don't understand well some details in the text ).

Thank you and best regards, Kostas vittas.

PS. Thanks to Darij for the editing of my post and sorry for my bad English.

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#6 h Jun 21, 2007, 5:47 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Quote:

Jacobi theorem. Let $\triangle ABC$ and let $X$ , $Y$ , $Z$ be three points such that $\{\begin{array}{c}m(\widehat{YAC})=m(\widehat{BAZ})=x\\\\ m(\widehat{ZBA})=m(\widehat{CBX})=y\\\\ m(\widehat{XCB})=m(\widehat{ACY})=z\end{array}$ .
Then the lines $AX$ , $BY$ , $CZ$ are concurrently.

The shortest proof. $\{\begin{array}{c}D\in AX\cap BC\\\ E\in BY\cap CA\\\ F\in CZ\cap AB\end{array}\|$ $\implies$ $\{\begin{array}{c}\frac{XD}{XA}=\frac{BD}{BA}\cdot\frac{\sin\widehat{XBD}}{\sin\widehat{XBA}}\\\\ \frac{XD}{XA}=\frac{CD}{CA}\cdot\frac{\sin\widehat{XCD}}{\sin\widehat{XCA}}\end{array}\|$ $\implies$ $\{\begin{array}{c}\frac{XD}{XA}=\frac{DB}{c}\cdot\frac{\sin y}{\sin (B+y)}\\\\ \frac{XD}{XA}=\frac{DC}{b}\cdot\frac{\sin z}{\sin (C+z)}\end{array}\|$ $\implies$

$\boxed{\frac{DB}{DC}=\frac{c}{b}\cdot\frac{\sin z}{\sin y}\cdot\frac{\sin (B+y)}{\sin (C+z)}}$ a.s.o. $\implies$ the lines $AX$ , $BY$ , $CZ$ are concurrently.

vittasko wrote:

A triangle $\bigtriangleup ABC$ is given, and let the external angle bisector of the angle $\angle A$ intersect the lines perpendicular to $BC$ and passing through $B$ and $C$ at the points $D$ and $E$ , respectively. Prove that the line segments $BE$ , $CD$ , $AO$ are concurrent, where $O$ is the circumcenter of $\bigtriangleup ABC$ .

Proof. Suppose w.l.o.g. that $B<C$ , i.e. $b<c$ . Denote the intersection $X\in AO\CD$. Observe that $AO\cap BE\cap CD\ne\emptyset$ $\Longleftrightarrow$ $\frac{XD}{XC}=\frac{BD}{CE}$ .

Thus, $m(\widehat{XAD})=m(\widehat{XAB})+m(\widehat{BAD})=(90-C)+(90-\frac{A}{2})=180-(C+\frac{A}{2})$ , $m(\widehat{XAC})=m(\widehat{OAC})=90-B$ and $\{\begin{array}{c}m(\widehat{ADB})=B+\frac{A}{2}\\\ m(\widehat{AEC})=C+\frac{A}{2}\end{array}$ .

Therefore, $\frac{XD}{XC}=$ $\frac{AD}{AC}\cdot\frac{\sin \widehat{XAD}}{\sin\widehat{XAC}}=$ $\frac{BD}{CE}\cdot \frac{AD}{BD}\cdot\frac{CE}{AC}\cdot\frac{\sin\widehat{XAD}}{\sin\widehat{XAC}}=$ $\frac{BD}{CE}\cdot\frac{\sin\widehat{ABD}}{\sin\widehat{BAD}}\cdot\frac{\sin\widehat{CAE}}{\sin\widehat{AEC}}\cdot\frac{\sin\widehat{XAD}}{\sin\widehat{XAC}}$ $\implies$

$\frac{XD}{XC}=$ $\frac{BD}{CE}\cdot\frac{\cos B}{\cos\frac{A}{2}}\cdot\frac{\cos\frac{A}{2}}{\sin(C+\frac{A}{2})}\cdot\frac{\sin(C+\frac{A}{2})}{\cos B}$ $\implies$ $\frac{XD}{XC}=\frac{BD}{CE}$ $\implies$ $X\in BE\cap CD$ $\implies$ the lines $AO$ , $BE$ , $CD$ are concurrently.

This post has been edited 6 times. Last edited by Virgil Nicula, Jun 21, 2007, 8:31 AM

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#7 h Jun 21, 2007, 8:17 AM • 2 Y

Y by Adventure10, Mango247

Thank you dear Virgil for another solution ( probably not the sortest...

) , of jacobi theorem.

I think that it is a good idea to collect all the proofs of this powerfull theorem we have in mind, as a separate topic in the "Theorems and formulas" area.

Have a nice day, Kostas Vittas.

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#8 h Jun 21, 2007, 7:31 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Quote:

Jacobi's theorem. Let $\triangle ABC$ and let $X$ , $Y$ , $Z$ be three points such that $\{\begin{array}{c}m(\widehat{YAC})=m(\widehat{BAZ})=x\\\\ m(\widehat{ZBA})=m(\widehat{CBX})=y\\\\ m(\widehat{XCB})=m(\widehat{ACY})=z\end{array}$ .
Then the lines $AX$ , $BY$ , $CZ$ are concurrently.

The shortest proof. $\{\begin{array}{c}D\in AX\cap BC\\\ E\in BY\cap CA\\\ F\in CZ\cap AB\end{array}\|$ $\implies$ $\{\begin{array}{c}\frac{XD}{XA}=\frac{BD}{BA}\cdot\frac{\sin\widehat{XBD}}{\sin\widehat{XBA}}\\\\ \frac{XD}{XA}=\frac{CD}{CA}\cdot\frac{\sin\widehat{XCD}}{\sin\widehat{XCA}}\end{array}\|$ $\implies$ $\{\begin{array}{c}\frac{XD}{XA}=\frac{DB}{c}\cdot\frac{\sin y}{\sin (B+y)}\\\\ \frac{XD}{XA}=\frac{DC}{b}\cdot\frac{\sin z}{\sin (C+z)}\end{array}\|$ $\implies$

$\boxed{\frac{DB}{DC}=\frac{c}{b}\cdot\frac{\sin z}{\sin y}\cdot\frac{\sin (B+y)}{\sin (C+z)}}$ a.s.o. $\implies$ the lines $AX$ , $BY$ , $CZ$ are concurrently.

vittasko wrote:

A triangle $\bigtriangleup ABC$ is given, and let the external angle bisector of the angle $\angle A$ intersect the lines perpendicular to $BC$ and passing through $B$ and $C$ at the points $D$ and $E$ , respectively. Prove that the line segments $BE$ , $CD$ , $AO$ are concurrent, where $O$ is the circumcenter of $\bigtriangleup ABC$ .

Proof I. Suppose w.l.o.g. that $B<C$ , i.e. $b<c$ . Denote the intersection $X\in AO\CD$. Observe that $AO\cap BE\cap CD\ne\emptyset$ $\Longleftrightarrow$ $\frac{XD}{XC}=\frac{BD}{CE}$ .

Thus, $m(\widehat{XAD})=m(\widehat{XAB})+m(\widehat{BAD})=(90-C)+(90-\frac{A}{2})=180-(C+\frac{A}{2})$ , $m(\widehat{XAC})=m(\widehat{OAC})=90-B$ and $\{\begin{array}{c}m(\widehat{ADB})=B+\frac{A}{2}\\\ m(\widehat{AEC})=C+\frac{A}{2}\end{array}$ .

Therefore, $\frac{XD}{XC}=$ $\frac{AD}{AC}\cdot\frac{\sin \widehat{XAD}}{\sin\widehat{XAC}}=$ $\frac{BD}{CE}\cdot \frac{AD}{BD}\cdot\frac{CE}{AC}\cdot\frac{\sin\widehat{XAD}}{\sin\widehat{XAC}}=$ $\frac{BD}{CE}\cdot\frac{\sin\widehat{ABD}}{\sin\widehat{BAD}}\cdot\frac{\sin\widehat{CAE}}{\sin\widehat{AEC}}\cdot\frac{\sin\widehat{XAD}}{\sin\widehat{XAC}}$ $\implies$

$\frac{XD}{XC}=$ $\frac{BD}{CE}\cdot\frac{\cos B}{\cos\frac{A}{2}}\cdot\frac{\cos\frac{A}{2}}{\sin(C+\frac{A}{2})}\cdot\frac{\sin(C+\frac{A}{2})}{\cos B}$ $\implies$ $\frac{XD}{XC}=\frac{BD}{CE}$ $\implies$ $X\in BE\cap CD$ $\implies$ the lines $AO$ , $BE$ , $CD$ are concurrently.

Proof II. Suppose w.l.o.g. that $B<C$ and denote the intersections $\{\begin{array}{c}X\in AO\cap CD\\\ P\in AO\ ,\ BP\parallel DE\end{array}$ . Prove easily that $\widehat{ADB}\equiv \widehat{DAP}$ , i.e. $PA=BD$ .

Observe that $\triangle BDP\equiv\triangle ECA$ . Therefore, $\frac{XD}{XC}=\frac{DP}{CA}=\frac{BD}{EC}$ , i.e. $X\in BE\cap CD$ .

vittasko wrote:

Thank you dear Virgil for another solution ( probably not the shortest... ) , of the Jacobi's theorem.

Vittasko, I "pester" you again : is shortestly the last my proof ? I am an wag and you are the sense of humour ...

Quote:

Some reciprocal questions. Let $BCED$ be a trapezoid for which $BD\parallel CE$ and $BD\perp BC$ . Consider the points : $\{\begin{array}{c}M\in (DE)\ ,\ N\in (BC)\ ;\ CM\parallel DN\\\ S\in (BC)\ ,\ SM\perp DE\\\ P\in DN\ ,\ BP\parallel DE\\\ X\in MP\cap DC\end{array}$ .
Prove the following the chain of the equivalencies : $BD=PM\Longleftrightarrow \widehat{BMS}\equiv\widehat{CMS}\Longleftrightarrow X\in BE\cap CD$ .

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#9 h Jun 22, 2007, 7:24 PM • 7 Y

Y by futurestar, tutubixu9198, Adventure10, GeoKing, and 3 other users

Dear all my friends,

I would like to presend a proof of Jacobi theorem, in its more general formulation as follows:

A GENERALIZATION OF JACOBI THEOREM.

ISOGONIC THEOREM. – Through vertices of a given triangle, we draw two lines isogonal conjugates with respect to the corresponded angle. Prove that the lines connecting each vertex of the given triangle with the intersection point of the isogonal lines through the other vertices, are concurrent at one point.

PROOF. ( The simplest I have in mind ).

We define three points $X,$ $Y,$ $Z,$ such that $\angle BXC =\angle x,$ $\angle AYC =\angle y,$ $\angle AZB =\angle z$ and we denote as $(O_{1}),$ $(O_{2}),$ $(O_{3}),$ the circumcircles of the triangles $\bigtriangleup XBC,$ $\bigtriangleup YAC,$ $\bigtriangleup ZAB,$ respectively.

Let $A''$ be, the intersection point of $AA',$ from the circumcircle of the triangle $\bigtriangleup A'BC$ and it is easy to show that $\angle A'A''C =\angle A'BC =\angle y$ and $\angle A'A''B =\angle A'CB =\angle z$ $,(1)$

From $(1)$ $\Longrightarrow$ $A'',$ lies on $(O_{2})$ and $(O_{3})$ and so, we have that $AA'\equiv AA'',$ is the radical axis of $(O_{2}),$ $(O_{3})$ and similarly for $BB'$ $($ radical axis of $(O_{1}),$ $(O_{3})$ $)$ and $CC'$ $($ radical axis of $(O_{1}),$ $(O_{2})$ $).$

Hence, we conclude that $AA'\cap BB'\cap CC'\equiv P,$ as the radical center of $(O_{1}),$ $(O_{2}),$ $(O_{3})$ and the proof is completed.

$\bullet$ This proof is dedicated to the memory of Theofilos Hrisostomidis.

REMARK. - An interest result of this powerful theorem is that it is true in every configuration of the isogonal lines through vertices of the given triangle $($ all of them inwardly or outwardly to $\bigtriangleup ABC,$ or some of them inwadrly and the other outwardly $).$

Also never mind how they have been taken, the intersection point of hte isogonal lines through vertices of the opposite side segment of the given triangle.

Kostas Vittas.

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jayme

#10 h Mar 3, 2010, 11:02 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
an article entitled "The Jacobi's theorem" has been put on my website.
http://perso.orange.fr/jl.ayme vol. 5
The originality of this article consists to prove this result by applying the Hesse's theorem then the desmic theorem. Many applications are presented.
Sincerely
Jean-Louis

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#11 h Mar 13, 2010, 6:54 PM • 1 Y

Y by Adventure10

how you all make these figures?what is the type of file called?where can i download it?

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#12 h Jul 17, 2010, 3:10 PM • 2 Y

Y by Adventure10, Mango247

If $\{D\} \equiv AX\cap BC$ , similarly getting $E \in AC$ and $F\in AB$ , then $\frac{BD}{CD}=\frac{[ABX]}{[ACX]}$ , where $[ABX]$ means the area of the $\triangle ABX$ , so $\frac{BD}{CD}=\frac{AB\cdot BX\cdot sin (B-y)}{AC\cdot CX \cdot sin (C-z)}$ $(1)$ and other two similar relations for vertices $B$ and $C$ respectively.
From the triangle $\triangle BCX$ we get $\frac{BX}{CX}=\frac{sin(z)}{sin(y)}$ $(2)$ and other two similar relations for the triangles $\triangle CAY$ and $\triangle ABZ$ .
Multiplying the three relations $(1)$ and taking into account the three relations $(2)$ we get $\frac{AF}{BF}\cdot \frac{BD}{CD}\cdot \frac{CE}{AE}=1$ , and the three lines $AX, BY, CZ$ are concurrent.

Remark: if the points $X$ , $Y$ or $Z$ are outside the given triangle, then the angles $x,y,z$ will be negative.

Best regards,
sunken rock

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swaqar

#13 h Jul 29, 2010, 7:59 PM • 2 Y

Y by Adventure10, Mango247

jayme wrote:

Dear Mathlinkers,
an article entitled "The Jacobi's theorem" has been put on my website.
http://perso.orange.fr/jl.ayme vol. 5
The originality of this article consists to prove this result by applying the Hesse's theorem then the desmic theorem. Many applications are presented.
Sincerely
Jean-Louis

Hey jayme please look here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=357322 and help . i need pdf and english version copies of those files and theorems on your website.

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