Обсуждение участника:Ahiskali

Теорема ЧивыПравить

$A{A}_1$$AA_{1}$ 
$B{B}_1$$BB_{1}$ 
$C{C}_1$$CC_{1}$ 
$\mathrm{△<!--\; △\; -->}AB{B}_1$$\triangle ABB_{1}$ 
$\frac{\stackrel{\to <!--\; \to \; -->}{A{C}_1}}{\stackrel{\to <!--\; \to \; -->}{C_{1}B}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{BO}}{\stackrel{\to <!--\; \to \; -->}{O{B}_1}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{B_{1}C}}{\stackrel{\to <!--\; \to \; -->}{OA}}=-<!--\; -\; -->1$${\frac {\overrightarrow {AC_{1}}}{\overrightarrow {C_{1}B}}}\cdot {\frac {\overrightarrow {BO}}{\overrightarrow {OB_{1}}}}\cdot {\frac {\overrightarrow {B_{1}C}}{\overrightarrow {OA}}}=-1$ 
$\mathrm{△<!--\; △\; -->}BC{B}_1$$\triangle BCB_{1}$ 

$\frac{\stackrel{\to <!--\; \to \; -->}{B_{1}O}}{\stackrel{\to <!--\; \to \; -->}{OB}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{B{A}_1}}{\stackrel{\to <!--\; \to \; -->}{A_{1}C}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{CA}}{\stackrel{\to <!--\; \to \; -->}{A{B}_1}}=-<!--\; -\; -->1$${\frac {\overrightarrow {B_{1}O}}{\overrightarrow {OB}}}\cdot {\frac {\overrightarrow {BA_{1}}}{\overrightarrow {A_{1}C}}}\cdot {\frac {\overrightarrow {CA}}{\overrightarrow {AB_{1}}}}=-1$ 

$\left(\frac{\stackrel{\to <!--\; \to \; -->}{A{C}_1}}{\stackrel{\to <!--\; \to \; -->}{C_{1}B}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{BO}}{\stackrel{\to <!--\; \to \; -->}{O{B}_1}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{B_{1}O}}{\stackrel{\to <!--\; \to \; -->}{OA}}\right)\cdot <!--\; \cdot \; -->\left(\frac{\stackrel{\to <!--\; \to \; -->}{B_{1}O}}{\stackrel{\to <!--\; \to \; -->}{OB}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{B{A}_1}}{\stackrel{\to <!--\; \to \; -->}{A_{1}O}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{CA}}{\stackrel{\to <!--\; \to \; -->}{A{B}_1}}\right)=(-<!--\; -\; -->1)\cdot <!--\; \cdot \; -->(-<!--\; -\; -->1)$$\left({\frac {\overrightarrow {AC_{1}}}{\overrightarrow {C_{1}B}}}\cdot {\frac {\overrightarrow {BO}}{\overrightarrow {OB_{1}}}}\cdot {\frac {\overrightarrow {B_{1}O}}{\overrightarrow {OA}}}\right)\cdot \left({\frac {\overrightarrow {B_{1}O}}{\overrightarrow {OB}}}\cdot {\frac {\overrightarrow {BA_{1}}}{\overrightarrow {A_{1}O}}}\cdot {\frac {\overrightarrow {CA}}{\overrightarrow {AB_{1}}}}\right)=(-1)\cdot (-1)$ 

$\frac{\stackrel{\to <!--\; \to \; -->}{A{C}_1}}{\stackrel{\to <!--\; \to \; -->}{C_{1}B}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{B{A}_1}}{\stackrel{\to <!--\; \to \; -->}{A_{1}C}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{C{B}_1}}{\stackrel{\to <!--\; \to \; -->}{B_{1}A}}=1$${\frac {\overrightarrow {AC_{1}}}{\overrightarrow {C_{1}B}}}\cdot {\frac {\overrightarrow {BA_{1}}}{\overrightarrow {A_{1}C}}}\cdot {\frac {\overrightarrow {CB_{1}}}{\overrightarrow {B_{1}A}}}=1$ 

Теорема МенелаяПравить

$\frac{\stackrel{\to <!--\; \to \; -->}{A{C}_1}}{\stackrel{\to <!--\; \to \; -->}{C_{1}B}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{B{A}_1}}{\stackrel{\to <!--\; \to \; -->}{A_{1}C}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{C{B}_1}}{\stackrel{\to <!--\; \to \; -->}{B_{1}A}}=-<!--\; -\; -->1$${\frac {\overrightarrow {AC_{1}}}{\overrightarrow {C_{1}B}}}\cdot {\frac {\overrightarrow {BA_{1}}}{\overrightarrow {A_{1}C}}}\cdot {\frac {\overrightarrow {CB_{1}}}{\overrightarrow {B_{1}A}}}=-1$ 

$\left.{\begin{matrix}{\overrightarrow {AC_{1}}}\uparrow \uparrow {\overrightarrow {C_{1}B}}\to {\frac {\overrightarrow {AC_{1}}}{\overrightarrow {C_{1}B}}}>0\\{\overrightarrow {BA_{1}}}\uparrow \uparrow {\overrightarrow {A_{1}C}}\to {\frac {\overrightarrow {BA_{1}}}{\overrightarrow {A_{1}C}}}>0\\{\overrightarrow {CB_{1}}}\uparrow \downarrow {\overrightarrow {B_{1}A}}\to {\frac {\overrightarrow {CB_{1}}}{\overrightarrow {B_{1}A}}}>0\end{matrix}}\right\}{\frac {\overrightarrow {AC_{1}}}{\overrightarrow {C_{1}B}}}\cdot {\frac {\overrightarrow {BA_{1}'}}{\overrightarrow {A'_{1}C}}}\cdot {\frac {\overrightarrow {CB_{1}}}{\overrightarrow {B_{1}A}}}<0$ 
$\mathrm{△<!--\; △\; -->}AK{C}_1\sim <!--\; \sim \; -->\mathrm{△<!--\; △\; -->}BM{C}_1$$\triangle AKC_{1}\sim \triangle BMC_{1}$ 
$\frac{A{C}_1}{B{C}_1}=\frac{AK}{BM}$${\frac {AC_{1}}{BC_{1}}}={\frac {AK}{BM}}$ 

$\mathrm{△<!--\; △\; -->}BM{A}_1\sim <!--\; \sim \; -->\mathrm{△<!--\; △\; -->}CN{A}_1$$\triangle BMA_{1}\sim \triangle CNA_{1}$ 
$\frac{B{A}_1}{C{A}_1}=\frac{BM}{CN}$${\frac {BA_{1}}{CA_{1}}}={\frac {BM}{CN}}$ 

$\mathrm{△<!--\; △\; -->}{B}_{1}CN\sim <!--\; \sim \; -->\mathrm{△<!--\; △\; -->}{B}_{1}AK$$\triangle B_{1}CN\sim \triangle B_{1}AK$ 
$\frac{B_{1}C}{B{1}_A}=\frac{CN}{AK}$${\frac {B_{1}C}{B1_{A}}}={\frac {CN}{AK}}$ 

$(1)$$(1)$ 
$(1),(2)\to <!--\; \to \; -->\frac{\stackrel{\to <!--\; \to \; -->}{B{A}_1}}{\stackrel{\to <!--\; \to \; -->}{A_{1}C}}=\frac{\stackrel{\to <!--\; \to \; -->}{B{A}_1}}{\stackrel{\to <!--\; \to \; -->}{A_{1}C}}$$(1),(2)\to {\frac {\overrightarrow {BA_{1}}}{\overrightarrow {A_{1}C}}}={\frac {\overrightarrow {BA_{1}}}{\overrightarrow {A_{1}C}}}$ 

$\frac{\stackrel{\to <!--\; \to \; -->}{AK}}{\stackrel{\to <!--\; \to \; -->}{KD}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{DO}}{\stackrel{\to <!--\; \to \; -->}{OB}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{BM}}{\stackrel{\to <!--\; \to \; -->}{MA}}=1\cdot <!--\; \cdot \; -->\frac{AB}{BC}\cdot <!--\; \cdot \; -->\left(\frac{BC}{AD}\right)=(-<!--\; -\; -->1)$${\frac {\overrightarrow {AK}}{\overrightarrow {KD}}}\cdot {\frac {\overrightarrow {DO}}{\overrightarrow {OB}}}\cdot {\frac {\overrightarrow {BM}}{\overrightarrow {MA}}}=1\cdot {\frac {AB}{BC}}\cdot \left({\frac {BC}{AD}}\right)=(-1)$ 

$\frac{\stackrel{\to <!--\; \to \; -->}{DO}}{\stackrel{\to <!--\; \to \; -->}{OB}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{BN}}{\stackrel{\to <!--\; \to \; -->}{NC}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{CM}}{\stackrel{\to <!--\; \to \; -->}{MD}}=\frac{AB}{BC}\cdot <!--\; \cdot \; -->1\cdot <!--\; \cdot \; -->\left(-<!--\; -\; -->\frac{BC}{AD}\right)=(-<!--\; -\; -->1)$${\frac {\overrightarrow {DO}}{\overrightarrow {OB}}}\cdot {\frac {\overrightarrow {BN}}{\overrightarrow {NC}}}\cdot {\frac {\overrightarrow {CM}}{\overrightarrow {MD}}}={\frac {AB}{BC}}\cdot 1\cdot \left(-{\frac {BC}{AD}}\right)=(-1)$ 

$B{B}_2((\cdot <!--\; \cdot \; -->)B_2\mathrm{ϵ<!--\; ϵ\; -->}[AC])$$BB_{2}((\cdot )B_{2}\epsilon [AC])$ 
$(\cdot <!--\; \cdot \; -->)O$$(\cdot )O$ 

$\frac{\stackrel{\to <!--\; \to \; -->}{C{B}_1}}{\stackrel{\to <!--\; \to \; -->}{B_{1}A}}=\frac{\stackrel{\to <!--\; \to \; -->}{C{B}_2}}{\stackrel{\to <!--\; \to \; -->}{B_{2}A}}$${\frac {\overrightarrow {CB_{1}}}{\overrightarrow {B_{1}A}}}={\frac {\overrightarrow {CB_{2}}}{\overrightarrow {B_{2}A}}}$ 

$\frac{\stackrel{\to <!--\; \to \; -->}{A{C}_1}}{\stackrel{\to <!--\; \to \; -->}{C_{1}B}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{B{A}_1}}{\stackrel{\to <!--\; \to \; -->}{A_{1}C}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{C{B}_1}}{\stackrel{\to <!--\; \to \; -->}{B_{1}A}}=1\cdot <!--\; \cdot \; -->1\cdot <!--\; \cdot \; -->1=1$${\frac {\overrightarrow {AC_{1}}}{\overrightarrow {C_{1}B}}}\cdot {\frac {\overrightarrow {BA_{1}}}{\overrightarrow {A_{1}C}}}\cdot {\frac {\overrightarrow {CB_{1}}}{\overrightarrow {B_{1}A}}}=1\cdot 1\cdot 1=1$ 

$\frac{\stackrel{\to <!--\; \to \; -->}{A{C}_1}}{\stackrel{\to <!--\; \to \; -->}{C_{1}B}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{B{A}_1}}{\stackrel{\to <!--\; \to \; -->}{A_{1}C}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{C{B}_1}}{\stackrel{\to <!--\; \to \; -->}{B_{1}A}}=\frac{A{C}_1}{C_{1}B}\cdot <!--\; \cdot \; -->\frac{B{A}_1}{A_{1}C}\cdot <!--\; \cdot \; -->\frac{C{B}_1}{B_{1}A}=\frac{AC}{AB}\cdot <!--\; \cdot \; -->\frac{AB}{AC}\cdot <!--\; \cdot \; -->\frac{BC}{AB}=1$${\frac {\overrightarrow {AC_{1}}}{\overrightarrow {C_{1}B}}}\cdot {\frac {\overrightarrow {BA_{1}}}{\overrightarrow {A_{1}C}}}\cdot {\frac {\overrightarrow {CB_{1}}}{\overrightarrow {B_{1}A}}}={\frac {AC_{1}}{C_{1}B}}\cdot {\frac {BA_{1}}{A_{1}C}}\cdot {\frac {CB_{1}}{B_{1}A}}={\frac {AC}{AB}}\cdot {\frac {AB}{AC}}\cdot {\frac {BC}{AB}}=1$ 

$\frac{\stackrel{\to <!--\; \to \; -->}{A{C}_1}}{\stackrel{\to <!--\; \to \; -->}{C_{1}B}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{B{A}_1}}{\stackrel{\to <!--\; \to \; -->}{A_{1}C}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{C{B}_1}}{\stackrel{\to <!--\; \to \; -->}{B_{1}A}}=\frac{A{C}_1}{C_{1}B}\cdot <!--\; \cdot \; -->\frac{B{A}_1}{A_{1}C}\cdot <!--\; \cdot \; -->\frac{C{B}_1}{B_{1}A}=\frac{AC}{AB}\cdot <!--\; \cdot \; -->\frac{AB}{AC}\cdot <!--\; \cdot \; -->\frac{BC}{AB}=1$${\frac {\overrightarrow {AC_{1}}}{\overrightarrow {C_{1}B}}}\cdot {\frac {\overrightarrow {BA_{1}}}{\overrightarrow {A_{1}C}}}\cdot {\frac {\overrightarrow {CB_{1}}}{\overrightarrow {B_{1}A}}}={\frac {AC_{1}}{C_{1}B}}\cdot {\frac {BA_{1}}{A_{1}C}}\cdot {\frac {CB_{1}}{B_{1}A}}={\frac {AC}{AB}}\cdot {\frac {AB}{AC}}\cdot {\frac {BC}{AB}}=1$ 

$\frac{\stackrel{\to <!--\; \to \; -->}{A{C}_1}}{\stackrel{\to <!--\; \to \; -->}{C_{1}B}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{B{A}_1}}{\stackrel{\to <!--\; \to \; -->}{A_{1}C}}\cdot <!--\; \cdot \; -->\frac{\stackrel{\to <!--\; \to \; -->}{C{B}_2}}{\stackrel{\to <!--\; \to \; -->}{B_{2}A}}=1$${\frac {\overrightarrow {AC_{1}}}{\overrightarrow {C_{1}B}}}\cdot {\frac {\overrightarrow {BA_{1}}}{\overrightarrow {A_{1}C}}}\cdot {\frac {\overrightarrow {CB_{2}}}{\overrightarrow {B_{2}A}}}=1$ 

Code, sweet codeПравить

$4S\mathrm{△<!--\; △\; -->}ABC$$4S\triangle ABC$ 
$\frac{BD}{AD}=\frac{BC}{AC}$${\frac {BD}{AD}}={\frac {BC}{AC}}$ 
$\mathrm{△<!--\; △\; -->}ABD$$\triangle ABD$ 
$\mathrm{△<!--\; △\; -->}ADC$$\triangle ADC$ 
$\frac{(1)}{(2)}:\frac{\frac{BD}{sin\mathrm{\alpha <!--\; \alpha \; -->}}=\frac{AB}{sin\mathrm{\beta <!--\; \beta \; -->}}}{\frac{CD}{sin\mathrm{\alpha <!--\; \alpha \; -->}}=\frac{AC}{sin\mathrm{\beta <!--\; \beta \; -->}}}$${\frac {(1)}{(2)}}:{\frac {{\frac {BD}{sin\alpha }}={\frac {AB}{sin\beta }}}{{\frac {CD}{sin\alpha }}={\frac {AC}{sin\beta }}}}$ 
$\frac{BD}{CD}=\frac{AB}{AC}$${\frac {BD}{CD}}={\frac {AB}{AC}}$ 
$\mathrm{\angle <!--\; \angle \; -->}AMD=\frac{\cup <!--\; \cup \; -->AD+\cup <!--\; \cup \; -->BC}{2}$$\angle AMD={\frac {\cup AD+\cup BC}{2}}$ 
$(\cdot <!--\; \cdot \; -->)O;R$$(\cdot )O;R$ 
$AC\cap <!--\; \cap \; -->BD=(\cdot <!--\; \cdot \; -->)M$$AC\cap BD=(\cdot )M$ 
$\mathrm{\angle <!--\; \angle \; -->}BDC=\frac{1}{2}\cup <!--\; \cup \; -->BC$$\angle BDC={\frac {1}{2}}\cup BC$ 
$\mathrm{\angle <!--\; \angle \; -->}ACD=\frac{1}{2}\cup <!--\; \cup \; -->AD$$\angle ACD={\frac {1}{2}}\cup AD$ 
$\mathrm{△<!--\; △\; -->}MDC$$\triangle MDC$ 
$\mathrm{\angle <!--\; \angle \; -->}AMD$$\angle AMD$ 
$AC=d,\mathrm{\angle <!--\; \angle \; -->}CAD=\mathrm{\alpha <!--\; \alpha \; -->}$$AC=d,\angle CAD=\alpha$ 
$S_{abcd}$$S_{abcd}$ 
$\mathrm{△<!--\; △\; -->}ACD(\mathrm{\angle <!--\; \angle \; -->}D={90}^{\circ <!--\; \circ \; -->})$$\triangle ACD(\angle D=90^{\circ })$ 
$\frac{CD}{AC}=sin\mathrm{\alpha <!--\; \alpha \; -->}\to <!--\; \to \; -->CD=d\cdot <!--\; \cdot \; -->sin\mathrm{\alpha <!--\; \alpha \; -->}$${\frac {CD}{AC}}=sin\alpha \to CD=d\cdot sin\alpha$ 
$\frac{AD}{AC}=cos\mathrm{\alpha <!--\; \alpha \; -->}\to <!--\; \to \; -->AD=d\cdot <!--\; \cdot \; -->cos\mathrm{\alpha <!--\; \alpha \; -->}$${\frac {AD}{AC}}=cos\alpha \to AD=d\cdot cos\alpha$ 
$S_{abcd}=AD\cdot <!--\; \cdot \; -->CD=d^2\cdot <!--\; \cdot \; -->sin\mathrm{\alpha <!--\; \alpha \; -->}\cdot <!--\; \cdot \; -->cos\mathrm{\alpha <!--\; \alpha \; -->}=\frac{d^2\cdot <!--\; \cdot \; -->sin2\mathrm{\alpha <!--\; \alpha \; -->}}{2}$$S_{abcd}=AD\cdot CD=d^{2}\cdot sin\alpha \cdot cos\alpha ={\frac {d^{2}\cdot sin2\alpha }{2}}$ 
$S_{abcd}=\frac{d^2\cdot <!--\; \cdot \; -->sin2\mathrm{\alpha <!--\; \alpha \; -->}}{2}$$S_{abcd}={\frac {d^{2}\cdot sin2\alpha }{2}}$