2019-I-14

$\begin{array}{rcll}
BC & = & CB & \text{(common side)} \\
\angle CBG & = & \angle BCF & \text{(alt. $\angle$s, $BG//EC$)} \\
\angle BCG & = & \angle CBF & \text{(alt. $\angle$s, $CG//DB$)} \\
\end{array}$

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

In $\Delta BCF$ and $\Delta DEF$,

$\begin{array}{rcll}
\angle BFC & = & \angle DFE & \text{(vert. opp. $\angle$s)} \\
\angle BCF & = & \angle DEF & \text{(alt. $\angle$s, $AD//BC$)} \\
\angle CBF & = & \angle EDF & \text{(alt. $\angle$s, $AD//BC$)}
\end{array}$

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

$\begin{array}{rcll}
\angle BFC & = & \angle CGB & \text{(corr. $\angle$s, $\cong \Delta$s)} \\
\angle BCF & = & \angle BGC & \text{(given)} \\
\angle BCF & = & \angle BFC \\
BF & = & BC & \text{(sides opp. eq. $\angle$s)}
\end{array}$

In $\Delta BCD$,

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

$\begin{array}{rcll}
\dfrac{BC}{DE} & = & \dfrac{BF}{DF} & \text{(corr. sides, $\sim \Delta$s)} \\
\dfrac{\ell}{DE} & = & \dfrac{\ell}{(\sqrt{2} – 1) \ell} \\
DE & = & (\sqrt{2} – 1) \ell
\end{array}$

Therefore, we have

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

By the result of (b)(i), we have

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

Therefore, I agree with the claim.