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Fie \( ABC \) un triunghi si punctele \( D\in (AC) \) , \( M\in (BD) \) .

Fie proprietatile : \( \left|\ \begin{array}{cc}
\mathrm{p\ :} & m(\angle {MCB})\ =\ m(\angle{MBA})\\\\\\
\mathrm{q\ :} & AM\perp MC\\\\\\
\mathrm{r\ :} & \frac {DA}{DC}=\frac {c^2\cdot \left(a^2+c^2-b^2\right)}{a^2\cdot \left(b^2+c^2\right)-\left(b^2-c^2\right)^2}\end{array}\right|\ . \) Sa se arate ca

\( (\ p\ \wedge\ q\ \rightarrow\ r\ )\ \ \wedge\ \ (\ p\ \wedge\ r\ \rightarrow\ q\ )\ \ \wedge\ \ (\ q\ \wedge\ r\ \rightarrow\ p\ )\ . \)

Cazuri particulare interesante : \( \left|\ \begin{array}{ccc}
b=c & \Longrightarrow & \frac {DA}{DC}=\frac 12\\\\\\
A=90^{\circ} & \Longrightarrow & \frac {DA}{DC}=\frac {c^2}{2b^2}\\\\\\
C=90^{\circ} & \Longrightarrow & \frac {DA}{DC}=\left(\frac cb\right)^2\\\\\\
B=60^{\circ} & \Longrightarrow & \frac {DA}{DC}=\frac {c^2}{a(a+c)}\end{array}\ \right|\ . \) Vezi aici / aici / aici cateva aplicatii.

Buna ziua!

Sa consideram urmatoarea problema:
In \( \Delta ABC \) cercul \( \Gamma \) de diametru \( AC \) intersecteaza din nou dreapta \( BC \) in punctul \( E \). Daca \( D\in\left(AC\right) \) si \( \left\{F\right\}=BD\cap\Gamma \), \( F\in Ext\left(\Delta ABC\right) \), atunci urmatoarele afirmatii sunt echivalente:
1. \( EF||AB \)
2. \( \frac{DA}{DC}=\frac{c^{2}\cdot\left(a^{2}+c^{2}-b^{2}\right)}{a^{2}\cdot\left(b^{2}+c^{2}\right)-\left(b^{2}-c^{2}\right)^{2}} \)

Solutie:
\( 1\Rightarrow 2 \)
\( \left\{N\right\}=EF\cap AC \) si notam cu \( S \) aria triunghiului \( \Delta ABC \).
\( S=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)} \), unde \( p=\frac{a+b+c}{2}\Rightarrow S=\frac{\sqrt{\left[\left(b+c\right)^{2}-a^{2}\right]\cdot\left[a^{2}-\left(b-c\right)^{2}\right]}}{4} \)
\( S=\frac{AE\cdot BC}{2}\Rightarrow AE=\frac{2S}{a} \)
\( CE^{2}=AC^{2}-AE^{2}=b^{2}-\frac{4S^{2}}{a^{2}}=\frac{a^{2}b^{2}-4S^{2}}{a^{2}} \)

Observatie:
\( a^{2}b^{2}-4S^{2}=a^{2}b^{2}-\frac{\left[\left(b+c\right)^{2}-a^{2}\right]\cdot\left[a^{2}-\left(b-c\right)^{2}\right]}{4}=\frac{4a^{2}b^{2}-a^{2}\cdot\left(b+c\right)^{2}+\left(b^{2}-c^{2}\right)^{2}+a^{4}-a^{2}\cdot\left(b-c\right)^{2}}{4}=\frac{a^{4}+2a^{2}\cdot\left(b^{2}-c^{2}\right)+\left(b^{2}-c^{2}\right)^{2}}{4}=\frac{\left(a^{2}+b^{2}-c^{2}\right)^{2}}{4}\Rightarrow a^{2}b^{2}-4S^{2}=\frac{\left(a^{2}+b^{2}-c^{2}\right)^{2}}{4} \)
Deci \( CE=\frac{a^{2}+b^{2}-c^{2}}{2a} \).

\( \frac{EN}{AB}=\frac{CE}{CB}\Rightarrow EN=\frac{c\cdot\left(a^{2}+b^{2}-c^{2}\right)}{2a^{2}} \)
\( \frac{CN}{CA}=\frac{CE}{CB}\Rightarrow CN=\frac{b\cdot\left(a^{2}+b^{2}-c^{2}\right)}{2a^{2}} \)
\( AN=AC-CN\Rightarrow AN=\frac{b\cdot\left(a^{2}+c^{2}-b^{2}\right)}{2a^{2}} \)
\( AN\cdot CN=EN\cdot FN\Rightarrow FN=\frac{b^{2}\cdot\left(a^{2}+c^{2}-b^{2}\right)}{2a^{2}c} \)

\( \frac{DN}{DA}=\frac{NF}{AB}\Rightarrow\frac{DN}{DA}=\frac{b^{2}\cdot\left(a^{2}+c^{2}-b^{2}\right)}{2a^{2}c^{2}}\Rightarrow\frac{AN}{DA}=\frac{b^{2}\cdot\left(a^{2}+c^{2}-b^{2}\right)+2a^{2}c^{2}}{2a^{2}c^{2}}\Rightarrow
DA=\frac{bc^{2}\cdot\left(a^{2}+c^{2}-b^{2}\right)}{b^{2}\cdot\left(a^{2}+c^{2}-b^{2}\right)+2a^{2}c^{2}}\Rightarrow \frac{DA}{AC}=\frac{c^{2}\cdot\left(a^{2}+c^{2}-b^{2}\right)}{b^{2}\cdot\left(a^{2}+c^{2}-b^{2}\right)+2a^{2}c^{2}}\Rightarrow\frac{DA}{DC}=\frac{c^{2}\cdot\left(a^{2}+c^{2}-b^{2}\right)}{a^{2}\cdot\left(b^{2}+c^{2}\right)-\left(b^{2}-c^{2}\right)^{2}} \)

\( 2\Rightarrow 1 \)
Fie \( F_{1}\in\Gamma \) astfel incat \( EF_{1}||AB \) si \( \left\{D_{1}\right\}=BF_{1}\cap AC \).
Din prima implicatie rezulta ca \( \frac{D_{1}A}{D_{1}C}=\frac{c^{2}\cdot\left(a^{2}+c^{2}-b^{2}\right)}{a^{2}\cdot\left(b^{2}+c^{2}\right)-\left(b^{2}-c^{2}\right)^{2}}\Rightarrow D_{1}=D \) si \( F_{1}=F \). Deci \( EF||AB \).


Sa demonstram problema propusa:
Fie \( \Gamma \) cercul de diametru \( AC \), \( \left\{C,E\right\}=BC\cap \Gamma \) si \( \left\{F\right\}=BD\cap\Gamma \), \( F\in Ext\left(\Delta ABC\right) \).

i) \( \left(p\wedge q\right)\rightarrow r \)
\( AM\bot MC\Rightarrow M\in\Gamma \)
\( m\left(\angle MFE\right)=m\left(\angle MCE\right)=m\left(\angle MBA\right)\Rightarrow EF||AB\Rightarrow\frac{DA}{DC}=\frac{c^{2}\cdot\left(a^{2}+c^{2}-b^{2}\right)}{a^{2}\cdot\left(b^{2}+c^{2}\right)-\left(b^{2}-c^{2}\right)^{2} \)

ii) \( \left(p\wedge r\right)\rightarrow q \)
\( \frac{DA}{DC}=\frac{c^{2}\cdot\left(a^{2}+c^{2}-b^{2}\right)}{a^{2}\cdot\left(b^{2}+c^{2}\right)-\left(b^{2}-c^{2}\right)^{2}}\Rightarrow EF||AB \)
\( m\left(\angle MFE\right)=m\left(\angle MBA\right)=m\left(\angle MCB\right)\Rightarrow M\in\Gamma\Rightarrow AM\bot MC \)

iii) \( \left(q\wedge r\right)\rightarrow p \)
\( AM\bot MC\Rightarrow M\in\Gamma \)
\( \frac{DA}{DC}=\frac{c^{2}\cdot\left(a^{2}+c^{2}-b^{2}\right)}{a^{2}\cdot\left(b^{2}+c^{2}\right)-\left(b^{2}-c^{2}\right)^{2}}\Rightarrow EF||AB \)
\( m\left(\angle MCB\right)=m\left(\angle MFE\right)=m\left(\angle MBA\right) \)

Multa sanatate,
Petrisor Neagoe

In loc de $$, trebuie pus \( la inceput iar la sfarsit [/ tex].

Pentru mai multe detalii aici si aici un program care schimba singur dolarii in \( . \) \)

Multumesc pentru ajutor, Laurian Filip.
Multa sanatate!