2019-I-14

$\begin{array}{rcll}
BC & = & CB & \text{(公共邊)} \\
\angle CBG & = & \angle BCF & \text{(錯角，$BG//EC$)} \\
\angle BCG & = & \angle CBF & \text{(錯角，$CG//DB$)} \\
\end{array}$

所以，$\Delta BCG \cong \Delta CBF\ \text{(A.S.A.)}$。

在 $\Delta BCF$ 及 $\Delta DEF$，

$\begin{array}{rcll}
\angle BFC & = & \angle DFE & \text{(對頂角)} \\
\angle BCF & = & \angle DEF & \text{(錯角，$AD//BC$)} \\
\angle CBF & = & \angle EDF & \text{(錯角，$AD//BC$)}
\end{array}$

所以，$\Delta BCF \sim \Delta DEF\ \text{(A.A.A.)}$。

$\begin{array}{rcll}
\angle BFC & = & \angle CGB & \text{($\cong \Delta$ 的對應角)} \\
\angle BCF & = & \angle BGC & \text{(已知)} \\
\angle BCF & = & \angle BFC \\
BF & = & BC & \text{(等角對等邊)}
\end{array}$

在 $\Delta BCD$，

$\begin{array}{rcll}
\angle BDC & = & 45^\circ & \text{(正方形性質)} \\
\sin \angle BDC & = & \dfrac{BC}{BD} \\
\sin 45^\circ & = & \dfrac{BC}{BF + DF} \\
\dfrac{1}{\sqrt{2}} & = & \dfrac{\ell}{\ell + DF} \\
\ell + DF & = & \sqrt{2} \ell \\
DF & = & (\sqrt{2} – 1) \ell
\end{array}$

$\begin{array}{rcll}
\dfrac{BC}{DE} & = & \dfrac{BF}{DF} & \text{($\sim \Delta$ 的對應邊)} \\
\dfrac{\ell}{DE} & = & \dfrac{\ell}{(\sqrt{2} – 1) \ell} \\
DE & = & (\sqrt{2} – 1) \ell
\end{array}$

所以，可得

$\begin{array}{rcl}
AE & = & AD – DE \\
& = & \ell – (\sqrt{2} – 1) \ell \\
& = & (2 – \sqrt{2}) \ell \\
& = & 0.585\ 786\ 437 \ell \\
\end{array}$

利用 (b)(i) 的結果，可得

$\begin{array}{rcl}
DF & = & (\sqrt{2} – 1) \ell \\
DF & = & 0.414\ 213\ 562 \ell \\
& < & 0.585\ 786\ 437 \ell \\ & = & AE \end{array}$

所以，我同意該宣稱。