ヘロンの公式の証明を5種類解説！中学生でも理解できる証明方法とは？

s=\frac{a+b+c}{2}

S=\sqrt{s(s-a)(s-b)(s-c)}  \\

s=\frac{3+7+8}{2}=9

\begin{align*}
S&=\sqrt{9(9-3)(9-7)(9-8)}  \\
&=\sqrt{9\cdot 6\cdot 2\cdot 1}  \\
&=6\sqrt{3}
\end{align*}

s=\frac{3+6+8}{2}=\frac{17}{2}

\begin{align*}
S&=\sqrt{\frac{17}{2}\left(\frac{17}{2}-4\right)\left(\frac{17}{2}-7\right)\left(\frac{17}{2}-8\right)}  \\
\\
&=\sqrt{\frac{17}{2} \cdot \frac{9}{2} \cdot \frac{3}{2} \cdot \frac{1}{2}}  \\
\\
&=\frac{3\sqrt{51}}{4}
\end{align*}

s=\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2}

\begin{align*}
S&=\sqrt{\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2}\left(\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2}-\sqrt{2}\right)\left(\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2}-\sqrt{3}\right)\left(\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2}-\sqrt{5}\right)}  \\
\\
&=\sqrt{\left(\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2}\right)\left(\frac{-\sqrt{2}+\sqrt{3}+\sqrt{5}}{2}\right)\left(\frac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{2}\right)\left(\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2}\right)}  \\
\\
&=\sqrt{\left(\frac{-2+3+2\sqrt{15}+5}{2^2}\right)\left(\frac{2-3+2\sqrt{15}-5}{2^2}\right)}  \\
\\
&=\sqrt{\frac{(2\sqrt{15}+6)(2\sqrt{15}-6)}{2^2\cdot 2^2}}  \\
\\
&=\sqrt{\frac{24}{2^2\cdot 2^2}}  \\
\\
&=\sqrt{\frac{6}{2^2}}  \\
\\
&=\frac{\sqrt{6}}{2}  \\
\end{align*}

AH^2=c^2-x^2~~~\cdots①

AH^2=b^2-(a-x)^2~~~\cdots②

\begin{align*}
c^2-x^2 &= b^2-(a-x)^2 \\
c^2-x^2 &= b^2-(a^2-2ax+x^2) \\
c^2-x^2 &= b^2-a^2+2ax-x^2 \\
-2ax&=b^2-a^2-c^2 \\
x&=\displaystyle \frac{a^2+c^2-b^2}{2a}
\end{align*}

\begin{align*}
AH&=\sqrt{c^2-x^2} \\
\\
&=\displaystyle \sqrt{c^2-\frac{(a^2+c^2-b^2)^2}{4 a^2}} \\
\\
&=\sqrt{c^2-\frac{a^4+c^4+b^4+2 a^{2} c^2-2 b^{2} c^2-2 a^{2} b^2}{4 a^2}} \\
\\
&=\sqrt{\frac{4a^2c^2-(a^4+c^4+b^4+2 a^{2} c^2-2 b^{2} c^2-2 a^{2} b^2)}{4 a^2}} \\
\\
&=\sqrt{\frac{4a^2c^2-a^4-c^4-b^4-2 a^{2} c^2+2 b^{2} c^2+2 a^{2} b^2)}{4 a^2}} \\
\\
&=\sqrt{\frac{-a^4-c^4-b^4+2 a^{2} c^2+2 b^{2} c^2+2 a^{2} b^2)}{4 a^2}} \\
\\
&=\sqrt{\frac{-(a^4+b^4+c^4-2 a^{2} c^2-2 b^{2} c^2-2 a^{2} b^2)}{4 a^2}} \\
\\
&=\sqrt{\frac{-{ (a^2-b^2-c^2)^2-4b^{2}c^2 }}{4 a^2}} \\
\\
&=\sqrt{\frac{-{ (a^2-b^2-c^2)+2bc}{ (a^2-b^2-c^2)-2bc}}{4 a^2}} \\
\\
&=\sqrt{\frac{-{ a^2-(b^2+c^2-2bc)}{ a^2-(b^2+c^2+2bc)}}{4 a^2}} \\
\\
&=\sqrt{\frac{-{ a^2-(b-c)^2}{ a^2-(b+c)^2}}{4 a^2}} \\
\\
&=\sqrt{\frac{-(a+b-c)(a-b+c)(a+b+c)(a-b-c)}{4 a^2}} \\
\\
&=\sqrt{\frac{-(a+b+c)(a-b-c)(a-b+c)(a+b-c)}{4 a^2}} \\
\\
&=\sqrt{\frac{(a+b+c)(b+c-a)(a-b+c)(a+b-c)}{4 a^2}} \\
\\
&=\sqrt{\frac{(a+b+c)}{2}\cdot \frac{(b+c-a)}{2}\cdot \frac{(a-b+c)}{2}
\cdot \frac{(a+b-c)}{2}\cdot \frac{4}{a^2}} \\
\\
&=\sqrt{\frac{(a+b+c)}{2}\cdot \frac{(a+b+c-2a)}{2}\cdot \frac{(a+b+c-2b)}{2}
\cdot \frac{(a+b+c-2c)}{2}\cdot \frac{4}{a^2}} \\
\\
&=\frac{2}{a} \sqrt{\left( \frac{a+b+c}{2} \right) \left( \frac{a+b+c}{2}-a \right) \left( \frac{a+b+c}{2}-b \right) \left( \frac{a+b+c}{2}-c \right) }
\\
\end{align*}

AH=\displaystyle \frac{2}{a} \sqrt{s(s-a)(s-b)(s-c)}

\begin{align*}
△ABC&=\displaystyle \frac{1}{2} a \cdot AH \\
\\
&=\frac{1}{2} a \cdot \frac{2}{a} \sqrt{s(s-a)(s-b)(s-c)} \\
\\
&=\sqrt{s(s-a)(s-b)(s-c)}
\end{align*}

S=\frac{r}{2}(a+b+c)

S=rs~~~\cdots ①

S = CG・OD

S = \sqrt{CG^2・OD^2}~~~\cdots②

∠COB + ∠CHB =180^{\circ}~~~\cdots ③

∠COB + ∠AOF =180^{\circ}~~~\cdots ④

 ∠CHB= ∠AOF

BC : BH= AF : FO = BG : OD \quad \text{（Gの設定と半径から）}

BC : BG = BH : OD~~~\cdots⑤

BH : DO = BI : DI~~~\cdots⑥

BC : BG = BI: DI

\begin{align*}
\frac{BC}{BG}&=\frac{BI}{DI}   \\
\\
\frac{BC}{BG}+1&=\frac{BI}{DI}+1   \\
\\
\frac{BC+BG}{BG}&=\frac{BI+DI}{DI}   \\
\\
\frac{CG}{BG}&=\frac{BD}{DI}   \\
\\
\frac{CG^2}{BG \cdot CG}&=\frac{BD\cdot DC}{DI\cdot DC} ~~~\cdots⑦  \\
\end{align*}

\begin{align*}
DI : DO &= DO: DC  \\
DI\cdot DC&=DO^2
\end{align*}

\begin{align*}
\frac{CG^2}{BG \cdot CG}&=\frac{BD\cdot DC}{DO^2}  \\
\\
CG^2 \cdot DO^2&=CG\cdot BG \cdot BD \cdot DC~~~\cdots ⑧
\end{align*}

\begin{align*}
S&= \sqrt{CG\cdot BG \cdot BD \cdot DC}  \\
&= \sqrt{s(s-a)(s-b)(s-c)}  
\end{align*}

\begin{equation*}
S = \frac{1}{2}ab \sin C
\end{equation*}

\begin{equation*}
S = \frac{1}{2}ab \sqrt{1 - \cos^2 C}
\end{equation*}

\begin{equation*}
S = \frac{1}{2}ab \sqrt{1 - \left(\frac{a^2 + b^2 - c^2}{2ab}\right)^2}
\end{equation*}

\begin{align*}
S &= \frac{1}{2}ab \sqrt{1 - \left(\frac{a^2 + b^2 - c^2}{2ab}\right)^2} \\
&=\frac{1}{2}ab \sqrt{\frac{(2ab)^2 - (a^2 + b^2 - c^2)^2}{(2ab)^2}} \\
&= \frac{1}{2}ab\cdot \frac{1}{2ab} \sqrt{(2ab)^2 - (a^2 + b^2 - c^2)^2}  ~~~\cdots(平方の差)\\
&= \frac{1}{4} \sqrt{(2ab + a^2 + b^2 - c^2)(2ab - a^2 - b^2 + c^2)} \\
&= \frac{1}{4} \sqrt{\{(a+b)^2 - c^2\}\{c^2 - (a-b)^2\}} ~~~\cdots(平方の差)\\
&= \frac{1}{4} \sqrt{(a+b+c)(a+b-c)(c-a+b)(c+a-b)} \\
&= \sqrt{\frac{(a+b+c)}{2} \frac{(-a+b+c)}{2} \frac{(a-b+c)}{2} \frac{(a+b-c)}{2}} \\
&=\sqrt{\left(\frac{a+b+c}{2}\right)\left(\frac{a+b+c}{2}-a\right)\left(\frac{a+b+c}{2}-b\right)\left(\frac{a+b+c}{2}-c\right)}\\
\end{align*}

S=\sqrt{s(s-a)(s-b)(s-c)}  \\

S=\frac{r}{2}(a+b+c)

S=rs~~~\cdots ①

\begin{cases}
&x+y=c  \\
&y+z=a  \\
&z+x=b
\end{cases}

\begin{align*}
2(x+y+z)&=a+b+c  \\
x+y+z&=\frac{a+b+c}{2}   \\
b+y&=s   \\
y&=s-b　~~~\cdots ②
\end{align*}

\begin{align*}
S&= \frac{1}{2}cR + \frac{1}{2}bR - \frac{1}{2}aR \\
&= \frac{b+c-a}{2}~R \\
&= \frac{b+c+a-2a}{2}~R \\
&= \left( \frac{b+c+a}{2}-a \right)R  \\
&=(s-a)R  ~~~\cdots ③
\end{align*}

\begin{cases}
&b+w=c+v  \\
&v+w=a  \\
\end{cases}

\begin{align*}
b-v&=c+v-a \\
-2v&=-a-b+c   \\
v&=\frac{a+b-c}{2}   \\
v&=\frac{a+b+c-2c}{2}  \\
v&=\frac{a+b+c}{2}-c  \\
v&=s-c　~~~\cdots ④
\end{align*}

\angle DBI=\angle  EBI  \\
\angle FBP=\angle  EBP  \\

\begin{align*}
\angle IBP&=\angle EBI + \angle EBP  \\
&=\frac{1}{2}\angle DBE + \frac{1}{2}\angle FBE  \\
&=\frac{1}{2}(\angle DBE + \angle FBE)  \\
&=\frac{1}{2} \cdot 180^{\circ}  \\
&=90^{\circ}
\end{align*}

\angle BDI=\angle BFP=90^{\circ}  \\
\angle DBI=90^{\circ}-\angle FBP=\angle FPB

\begin{align*}
\text{DI} : \text{FB} &= \text{BD} : \text{PF}  \\
r : (s-c) &= (s-b) : R  \\
rR&=(s-b)(s-c)  ~~~\cdots ⑤
\end{align*}

\begin{equation*}
S^2=s(s-a)rR
\end{equation*}

\begin{align*}
S^2&=s(s-a)(s-b)(s-c)  \\
S&=\sqrt{s(s-a)(s-b)(s-c)}
\end{align*}

s=\frac{a+b+c+d}{2}

S=\sqrt{(s-a)(s-b)(s-c)(s-d)}  \\

s=\frac{a+b+c+d}{2}~~~\cdots①

S=\sqrt{(s-a)(s-b)(s-c)(s-d)} ~~~\cdots②

s=\frac{a+b+c+0}{2}=\frac{a+b+c}{2}

S=\sqrt{(s-a)(s-b)(s-c)(s-0)}=\sqrt{s(s-a)(s-b)(s-c)}