78. Jakobi's theorem.

by Virgil Nicula, Aug 4, 2010, 3:19 AM

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

Proof. $\left\{\begin{array}{c}D\in AX\cap BC\\\ E\in BY\cap CA\\\ F\in CZ\cap AB\end{array}\right\|$ $\implies$ $\left\{\begin{array}{c}\frac{XD}{XA}=\frac{BD}{BA}\cdot\frac{\sin\widehat{XBD}}{\sin\widehat{XBA}}\\\\ \frac{XD}{XA}=\frac{CD}{CA}\cdot\frac{\sin\widehat{XCD}}{\sin\widehat{XCA}}\end{array}\right\|$ $\implies$ $\left\{\begin{array}{c}\frac{XD}{XA}=\frac{DB}{c}\cdot\frac{\sin y}{\sin (B+y)}\\\\ \frac{XD}{XA}=\frac{DC}{b}\cdot\frac{\sin z}{\sin (C+z)}\end{array}\right\|$ $\implies$ $\boxed{\frac{DB}{DC}=\frac{c}{b}\cdot\frac{\sin z}{\sin y}\cdot\frac{\sin (B+y)}{\sin (C+z)}}$ a.s.o. $\implies$ the lines $AX$, $BY$, $CZ$ are concurrently.
Quote:
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 1. Suppose w.l.o.g. that $B<C$, i.e. $b<c$. Denote the intersection $X\in AO\cap CD$. Observe that $AO\cap BE\cap CD\ne\emptyset$ $\Longleftrightarrow$ $\frac{XD}{XC}=\frac{BD}{CE}$. Thus, $m(\widehat{XAD})=m(\widehat{XAB})+m(\widehat{BAD})=(90-C)+\left(90-\frac{A}{2}\right)=180-\left(C+\frac{A}{2}\right)$ , $m(\widehat{XAC})=m(\widehat{OAC})=90-B$ and $\left\{\begin{array}{c}m(\widehat{ADB})=B+\frac{A}{2}\\\ m(\widehat{AEC})=C+\frac{A}{2}\end{array}\right\|$. Therefore, $\frac{XD}{XC}=$ $\frac{AD}{AC}\cdot\frac{\sin \widehat{XAD}}{\sin\widehat{XAC}}=$ $\frac{BD}{CE}\cdot \frac{AD}{BD}\cdot\frac{CE}{AC}\cdot\frac{\sin\widehat{XAD}}{\sin\widehat{XAC}}=$ $\frac{BD}{CE}\cdot\frac{\sin\widehat{ABD}}{\sin\widehat{BAD}}\cdot\frac{\sin\widehat{CAE}}{\sin\widehat{AEC}}\cdot\frac{\sin\widehat{XAD}}{\sin\widehat{XAC}}$ $\implies$ $\frac{XD}{XC}=$ $\frac{BD}{CE}\cdot\frac{\cos B}{\cos\frac{A}{2}}\cdot\frac{\cos\frac{A}{2}}{\sin\left(C+\frac{A}{2}\right)}\cdot\frac{\sin\left(C+\frac{A}{2}\right)}{\cos B}$ $\implies$ $\frac{XD}{XC}=\frac{BD}{CE}$ $\implies$ $X\in BE\cap CD$ $\implies$ the lines $AO$, $BE$, $CD$ are concurrently.

Proof 2. Suppose w.l.o.g. that $B<C$ and denote the intersections $\left\{\begin{array}{c}X\in AO\cap CD\\\ P\in AO\ ,\ BP\parallel DE\end{array}\right\|$. 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. $X\in BE\cap CD$.
Quote:
Some reciprocal questions. Let $BCED$ be a trapezoid for which $BD\parallel CE$ and $BD\perp BC$. Consider the points :$\left\{\begin{array}{c}M\in (DE)\ ,\ N\in (BC)\ ;\ CM\parallel DN\\\ S\in (BC)\ ,\ SM\perp DE\\\ P\in DN\ ,\ BP\parallel DE\\\ X\in MP\cap DC\end{array}\right\|$.
Prove that the following the chain of the equivalencies : $BD=PM\Longleftrightarrow \widehat{BMS}\equiv\widehat{CMS}\Longleftrightarrow X\in BE\cap CD$.
This post has been edited 2 times. Last edited by Virgil Nicula, Nov 23, 2015, 3:28 PM

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    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

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