Związek między długościami boków
trójkąta a długościami jego środkowych
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{:[a=(2)/(3)sqrt(2m_(b)^(2)+2m_(c)^(2)-m_(a)^(2))],[b=(2)/(3)sqrt(2m_(c)^(2)+2m_(a)^(2)-m_(b)^(2))],[c=(2)/(3)sqrt(2m_(b)^(2)+2m_(a)^(2)-m_(c)^(2))]:} \begin{aligned}
&a=\frac{2}{3} \sqrt{2 m_{b}^{2}+2 m_{c}^{2}-m_{a}^{2}} \\
&b=\frac{2}{3} \sqrt{2 m_{c}^{2}+2 m_{a}^{2}-m_{b}^{2}} \\
&c=\frac{2}{3} \sqrt{2 m_{b}^{2}+2 m_{a}^{2}-m_{c}^{2}}
\end{aligned} a = 2 3 2 m b 2 + 2 m c 2 − m a 2 b = 2 3 2 m c 2 + 2 m a 2 − m b 2 c = 2 3 2 m b 2 + 2 m a 2 − m c 2
Dowód:
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Z/_\EBC \mathrm{Z} \triangle E B C Z △ E B C i tw. kosinusów:
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a^(2)=m_(c)^(2)+((1)/(2)c)^(2)-2m_(c)(1)/(2)c cos alpha=>a^(2)=m_(c)^(2)+(1)/(4)c^(2)-m_(c)c cos alpha a^{2}=m_{c}^{2}+\left(\frac{1}{2} c\right)^{2}-2 m_{c} \frac{1}{2} c \cos \alpha \Rightarrow a^{2}=m_{c}^{2}+\frac{1}{4} c^{2}-m_{c} c \cos \alpha a 2 = m c 2 + ( 1 2 c ) 2 − 2 m c 1 2 c cos α ⇒ a 2 = m c 2 + 1 4 c 2 − m c c cos α Z
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((2)/(3)m_(b))^(2)=((1)/(3)m_(c))^(2)+((1)/(2)c)^(2)-2*(1)/(3)m_(c)(1)/(2)c cos alpha \left(\frac{2}{3} m_{b}\right)^{2}=\left(\frac{1}{3} m_{c}\right)^{2}+\left(\frac{1}{2} c\right)^{2}-2 \cdot \frac{1}{3} m_{c} \frac{1}{2} c \cos \alpha ( 2 3 m b ) 2 = ( 1 3 m c ) 2 + ( 1 2 c ) 2 − 2 ⋅ 1 3 m c 1 2 c cos α Z
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{:[((2)/(3)m_(a))^(2)=((1)/(3)m_(c))^(2)+((1)/(2)c)^(2)-2*(1)/(3)m_(c)*(1)/(2)c cos(180^(@)-alpha)],[((2)/(3)m_(a))^(2)=((1)/(3)m_(c))^(2)+((1)/(2)c)^(2)+2*(1)/(3)m_(c)*(1)/(2)c cos alpha]:} \begin{aligned}
&\left(\frac{2}{3} m_{a}\right)^{2}=\left(\frac{1}{3} m_{c}\right)^{2}+\left(\frac{1}{2} c\right)^{2}-2 \cdot \frac{1}{3} m_{c} \cdot \frac{1}{2} c \cos \left(180^{\circ}-\alpha\right) \\
&\left(\frac{2}{3} m_{a}\right)^{2}=\left(\frac{1}{3} m_{c}\right)^{2}+\left(\frac{1}{2} c\right)^{2}+2 \cdot \frac{1}{3} m_{c} \cdot \frac{1}{2} c \cos \alpha
\end{aligned} ( 2 3 m a ) 2 = ( 1 3 m c ) 2 + ( 1 2 c ) 2 − 2 ⋅ 1 3 m c ⋅ 1 2 c cos ( 180 ∘ − α ) ( 2 3 m a ) 2 = ( 1 3 m c ) 2 + ( 1 2 c ) 2 + 2 ⋅ 1 3 m c ⋅ 1 2 c cos α
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{:[{[((2)/(3)m_(b))^(2)=((1)/(3)m_(c))^(2)+((1)/(2)c)^(2)-(1)/(3)m_(c)c cos alpha],[+((2)/(3)m_(a))^(2)=((1)/(3)m_(c))^(2)+((1)/(2)c)^(2)+(1)/(3)m_(c)c cos alpha]:}],[((2)/(3)m_(b))^(2)+((2)/(3)m_(a))^(2)=2((1)/(3)m_(c))^(2)+2*((1)/(2)c)^(2)],[(4)/(9)m_(b)^(2)+(4)/(9)m_(a)^(2)=(2)/(g)m_(c)^(2)+(1)/(2)c^(2)=>(1)/(2)c^(2)=(4)/(9)m_(b)^(2)+(4)/(9)m_(a)^(2)-(2)/(9)m_(c)^(2)//*2=>c^(2)=(8)/(9)m_(b)^(2)+(8)/(9)m_(a)^(2)-(4)/(9)m_(c)^(2)=>],[=>c^(2)=(4)/(9)(2m_(b)^(2)+2m_(a)^(2)-m_(c)^(2))=>c=(2)/(3)sqrt(2m_(b)^(2)+2m_(a)^(2)-m_(c)^(2))]:} \begin{aligned}
& \left\{\begin{array}{l}\left(\frac{2}{3} m_{b}\right)^{2}=\left(\frac{1}{3} m_{c}\right)^{2}+\left(\frac{1}{2} c\right)^{2}-\frac{1}{3} m_{c} c \cos \alpha \\+\left(\frac{2}{3} m_{a}\right)^{2}=\left(\frac{1}{3} m_{c}\right)^{2}+\left(\frac{1}{2} c\right)^{2}+\frac{1}{3} m_{c} c \cos \alpha\end{array}\right. \\
& \left(\frac{2}{3} m_{b}\right)^{2}+\left(\frac{2}{3} m_{a}\right)^{2}=2\left(\frac{1}{3} m_{c}\right)^{2}+2 \cdot\left(\frac{1}{2} c\right)^{2} \\
& \frac{4}{9} m_{b}^{2}+\frac{4}{9} m_{a}^{2}=\frac{2}{g} m_{c}^{2}+\frac{1}{2} c^{2} \Rightarrow \frac{1}{2} c^{2}=\frac{4}{9} m_{b}^{2}+\frac{4}{9} m_{a}^{2}-\frac{2}{9} m_{c}^{2} / \cdot 2 \Rightarrow c^{2}=\frac{8}{9} m_{b}^{2}+\frac{8}{9} m_{a}^{2}-\frac{4}{9} m_{c}^{2} \Rightarrow \\
& \Rightarrow c^{2}=\frac{4}{9}\left(2 m_{b}^{2}+2 m_{a}^{2}-m_{c}^{2}\right) \Rightarrow c=\frac{2}{3} \sqrt{2 m_{b}^{2}+2 m_{a}^{2}-m_{c}^{2}}
\end{aligned} { ( 2 3 m b ) 2 = ( 1 3 m c ) 2 + ( 1 2 c ) 2 − 1 3 m c c cos α + ( 2 3 m a ) 2 = ( 1 3 m c ) 2 + ( 1 2 c ) 2 + 1 3 m c c cos α ( 2 3 m b ) 2 + ( 2 3 m a ) 2 = 2 ( 1 3 m c ) 2 + 2 ⋅ ( 1 2 c ) 2 4 9 m b 2 + 4 9 m a 2 = 2 g m c 2 + 1 2 c 2 ⇒ 1 2 c 2 = 4 9 m b 2 + 4 9 m a 2 − 2 9 m c 2 / ⋅ 2 ⇒ c 2 = 8 9 m b 2 + 8 9 m a 2 − 4 9 m c 2 ⇒ ⇒ c 2 = 4 9 ( 2 m b 2 + 2 m a 2 − m c 2 ) ⇒ c = 2 3 2 m b 2 + 2 m a 2 − m c 2 Analogicznie wyprowadzamy wzory dla boku
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Zatem boki trójkąta można wyrazić za pomocą środkowych trójkąta w następujący sposób
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{:[a=(2)/(3)sqrt(2m_(b)^(2)+2m_(c)^(2)-m_(a)^(2))],[b=(2)/(3)sqrt(2m_(c)^(2)+2m_(a)^(2)-m_(b)^(2))],[c=(2)/(3)sqrt(2m_(b)^(2)+2m_(a)^(2)-m_(c)^(2))]:} \begin{aligned}
&a=\frac{2}{3} \sqrt{2 m_{b}^{2}+2 m_{c}^{2}-m_{a}^{2}} \\
&b=\frac{2}{3} \sqrt{2 m_{c}^{2}+2 m_{a}^{2}-m_{b}^{2}} \\
&c=\frac{2}{3} \sqrt{2 m_{b}^{2}+2 m_{a}^{2}-m_{c}^{2}}
\end{aligned} a = 2 3 2 m b 2 + 2 m c 2 − m a 2 b = 2 3 2 m c 2 + 2 m a 2 − m b 2 c = 2 3 2 m b 2 + 2 m a 2 − m c 2
