Question

Fie triunghiul ABC cu laturile a, b, c, iar I centrul cercului înscris, M şi N astfel încât BM(vectorial)=m*MA(vectoial) ,CN(vectorial)=n*NA(vectorial). Să se arate că M, I, N sunt coliniare dacă şi numai dacă mb + nc = a.
Vreau o rezolvare cu TEOREMA TRANSVERSALEI,daca se poate! Multumesc mult!

Answer 1

$\displaystyle Solutie~(asumandu-mi~ca~m,n\ \textgreater \ 0). \\ \\ Avem~M \in (AB)~si~N \in (AC),~ \frac{MB}{MA}=m~si~ \frac{NC}{NA}=n. \\ \\ Fie~ \{S \}= IM \cap BC~si~ \{T \}=IN \cap BC. \\ \\ Din~teorema~lui~Menelaus~in~triunghiul~ABD~cu~transversala \\ \\ S-M-I,~avem~ \frac{SD}{SB} \cdot \frac{MB}{MA} \cdot \frac{IA}{ID}=1. \\ \\ Din~teorema~lui~Menelaus~in~triunghiul~ADC~cu~trasnversala \\ \\ T-I-N,~avem~ \frac{TC}{TD} \cdot \frac{ID}{IA} \cdot \frac{NA}{NC}=1.$

$\displaystyle \frac{IA}{ID}= \frac{b+c}{a}~(aplicatie~directa~a~teoremei~lui~Van~Aubel~;~rezulta \\ \\ totodata~din~aplicarea~teoremei~lui~Menelaus~in~ \Delta BEC~cu~ \\ \\ transversala~B-I-E,~unde~E~este~piciorul~bisectoarei~din~B). \\ \\ Obtinem~ \frac{SD}{SB}= \frac{a}{m(b+c)}~si~ \frac{TD}{TC}= \frac{a}{n(b+c)}. \\ \\ Deci~ \frac{SD}{BD}= \frac{a}{a-m(b+c)}~si~\frac{TD}{CD}= \frac{a}{a-n(b+c)}.$

$\displaystyle Din~teorema~bisectoarei~si~proportii~derivate~rezulta~rapid \\ \\ BD= \frac{ab}{b+c}~si~CD= \frac{ac}{b+c}. \\ \\ Prin~inlocuire~rezulta:~SD= \frac{a^2b}{(a-n(b+c))(b+c)}~si \\ \\ TD= \frac{a^2c}{(a-m(b+c))(b+c)}. \\ \\ M,I,N-coliniare \Leftrightarrow SD=TD \Leftrightarrow calcule \Leftrightarrow mb+nc=a.$

$\displaystyle * Daca~nu~exista~acea~restrictie,~atunci~se~lucreaza~cu~ rapoarte \\ \\ orientate,~se~exprima~\overrightarrow{IM}~in~functie~de~ \overrightarrow{IA}~si~\overrightarrow{IB},~iar~\overrightarrow{IN}~in \\ \\ functie~de~\overrightarrow{IA}~si~\overrightarrow{IC}.$

$\displaystyle Se~tine~cont~ca~\overrightarrow{XI}= \frac{a\overrightarrow{XA}+b \overrightarrow{XB}+c \overrightarrow{XC}}{a+b+c}~pentru~orice~punct~X \\ \\ din~plan,~de~unde~se~exprima~\overrightarrow{IC}~in~functie~de~\overrightarrow{IA}~si~\overrightarrow{IB}~si \\ \\ astfel~s-a~gasit~o~exprimare~de~tipul~\overrightarrow{IM}=u\overrightarrow{IA}+v\overrightarrow{IB}~si \\ \\ \overrightarrow{IN}=p\overrightarrow{IA}+q \overrightarrow{IB}. \\ \\ M,I,N-coliniare \Leftrightarrow \frac{u}{p}= \frac{v}{q}.$