,,Nemoguć” dokaz Pitagorina teorema

1. Uvod

2. Dokaz Calcee Johnson i Ne’Kiye Jackson

1. dokaz

2. dokaz

1. ${}$\begin{equation}
   P_{\triangle ABM}=\dfrac{\left|AB \right| \cdot\left| BM\right| }{2}=\dfrac{abc^2}{b^2-a^2}\tag{9}
   \label{p1}
   \end{equation}
2. ${}$\begin{equation}
   \begin{aligned}
   P_{\triangle ABM}&=P_{\triangle ABD}+ P_{\triangle BDE}+P_{\triangle DEF}+P_{\triangle EFG}+\dots\\
   &=ab+\dfrac{2a^3}{b}+\dfrac{2a^5}{b^2}+\dfrac{2a^7}{b^5}+\dots\\
   &=ab+\dfrac{2a^3}{b}\cdot\left(1+\dfrac{a^2}{b^2}+\dfrac{a^4}{b^4}+\dots \right) \\
   &=ab+\dfrac{2a^3}{b}\cdot\dfrac{1}{1-\dfrac{a^2}{b^2}}\\
   &=\dfrac{ab\left(a^2+b^2 \right) }{b^2-a^2}
   \end{aligned}\tag{10}
   \label{p2}
   \end{equation}

3. Još dva slična dokaza Pitagorina teorema

3.1. Prvi dokaz

1. ${}$\begin{equation}
   P_{\triangle CAB}=\dfrac{\left|AC \right| \cdot\left|CB \right| }{2}=\dfrac{a\cdot b}{2}\tag{18}
   \label{nacin1}
   \end{equation}
2. ${}$\begin{equation}
   \begin{aligned}
   P_{\triangle CAB}&= P_{\triangle CAC_1}+P_{\triangle C_1C_2C}+P_{\triangle C_2C_1C_3}+P_{\triangle C_2C_3C_4}+\dots\\
   &=\dfrac{\left| CC_1\right|\cdot \left|AC_1 \right| }{2}+\dfrac{\left| CC_2\right|\cdot \left|C_1C_2 \right| }{2} +\dfrac{\left| C_1C_3\right|\cdot \left|C_2C_3 \right| }{2}+\dfrac{\left| C_2C_4\right|\cdot \left|C_3C_4 \right| }{2}+\dots\\
   &=\dfrac{b^2\sin\alpha\cos\alpha}{2} + \dfrac{b^2\sin^3\alpha\cos\alpha}{2}+\dfrac{b^2\sin^5\alpha\cos\alpha}{2}+\dfrac{b^2\sin^7\alpha\cos\alpha}{2} +\dots\\
   &=\dfrac{b^2\sin\alpha\cos\alpha}{2}\cdot \left(1+\sin^2\alpha+\sin^4\alpha+\sin^6\alpha+\dots \right) \\
   &=\dfrac{b^2\sin\alpha\cos\alpha}{2\left( 1-\sin^2\alpha\right) }.
   \end{aligned}\tag{19}
   \label{nacin}
   \end{equation}

3.2. Drugi dokaz

1. \begin{equation}
   P_{\triangle ABC}=\dfrac{ab}{2}\tag{26}
   \label{1}
   \end{equation}
2. Odredimo za početak $\left| EF\right|.$ Kako je $\triangle ABC\sim\triangle AEF$, slijedi \begin{equation*} \dfrac{\left|EF \right| }{\left|AF \right| }=\dfrac{a}{b}\Longrightarrow \left|EF \right|=\dfrac{a}{2b}\cdot \left(b-\dfrac{a^2}{b} \right). \end{equation*} Sada možemo površinu trokuta $ABC$ odrediti i na sljedeći način: \begin{equation}\begin{aligned}
   P_{\triangle ABC}&=P_{\triangle BCD}+P_{\triangle BDE}+P_{\triangle DAE}\\
   &=\dfrac{\left|BC \right|\cdot \left|CD \right| }{2}+\dfrac{\left|BD \right|\cdot \left|DE \right| }{2}+\dfrac{\left|AD \right|\cdot \left|EF \right| }{2}\\
   &=\dfrac{a^3}{2b}+\dfrac{ac^2}{4b^2}\cdot\left(b-\dfrac{a^2}{b} \right) +\dfrac{a}{4b}\cdot\left(b-\dfrac{a^2}{b} \right)^2\\ &=\dfrac{a^3}{2b}+\dfrac{ac^2\cdot\left(b^2-a^2 \right) }{4b^3}+\dfrac{a\cdot\left(b^2-a^2 \right)^2}{4b^3}\\ &=\dfrac{2b^2a^3+ac^2\left(b^2-a^2 \right)+a\left(b^2-a^2 \right)^2 }{4b^3}\end{aligned}\tag{27}\label{2}
   \end{equation}
   Izjednačavanjem formula (\ref{1}) i (\ref{2}) te množenjem s $\dfrac{4b^3}{a}$ slijedi Pitagorin poučak, \begin{equation} \begin{aligned}
   2b^4&=2a^2b^2+c^2\left(b^2-a^2 \right) +\left(b^2-a^2 \right)^2\\
   2b^2\left(b^2-a^2 \right) &=c^2\left(b^2-a^2 \right) +\left(b^2-a^2 \right)^2\\
   2b^2 &=c^2 +b^2-a^2\\
   a^2+b^2&=c^2. \end{aligned} \end{equation}

Literatura

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