\begin{equation}延申\mathrm{AB}，讓\mathrm{BH}=\mathrm{BC}=\mathrm{a}並延申\mathrm{AC}，讓\mathrm{BD}\perp\mathrm{CH}\end{equation}
\begin{equation}則\angle\mathrm{DBH}=\angle\mathrm{DBC}\end{equation}
\begin{equation}\therefore\triangle\mathrm{DBC}\cong\triangle\mathrm{DBH}\Rightarrow\angle\mathrm{DHB}=\mathrm{90}^\circ，\mathrm{DC}=\mathrm{DH}=\mathrm{x}\end{equation}
\begin{equation}\because\triangle\mathrm{ABC}\sim\triangle\mathrm{ADH}\end{equation}
\begin{equation}\therefore\mathrm{b}:(\mathrm{c}+\mathrm{a})=\mathrm{a}:\mathrm{x}\end{equation}
\begin{equation}\Rightarrow\mathrm{a}^2=\mathrm{bx}-\mathrm{ac}\cdots\cdots(\mathrm{1})\end{equation}
\begin{equation}\therefore\mathrm{b}:(\mathrm{c}+\mathrm{a})=\mathrm{c}:(\mathrm{b}+\mathrm{x})\end{equation}
\begin{equation}\Rightarrow\mathrm{bx}-\mathrm{ac}=\mathrm{c}^2-\mathrm{b}^2\cdots\cdots(\mathrm{2})\end{equation}
\begin{equation}將(\mathrm{2})代入(\mathrm{1})可以即可得到\end{equation}
\begin{equation}\Rightarrow\mathrm{a}^2+\mathrm{b}^2=\mathrm{c}^2\end{equation}
Reference:
1. Elisha S. Loomis (1935): « The Pythagorean Proposition ». p. 26.