G1.4 In Figure 1 , if $A B \perp C D, F$ is the midpoint of $B E, \angle A=45^{\circ}$, $D F=3, B D=4$ and $A D=n$, find the value of $n$.
Let $G$ be the foot of perpendicular drawn from $E$ onto $C F$.
$$
\begin{array}{l}
\angle B F D=\angle C F E \\
B F=5 \\
E F=5 \\
\angle B D F=90^{\circ}=\angle E G F \\
\triangle B D F \cong \triangle E G F \\
\therefore F G=D F=3 \\
E G=4 \\
\angle A C D=45^{\circ}
\end{array}
$$
(vert. opp. $\angle$ s)
(Pythagoras' Theorem)
(Given $F$ is the midpoint)
(by construction)
(A.A.S.)
(corr. sides, $\cong \Delta$ 's)
(corr. sides, $\cong \Delta$ 's)
( $\angle \mathrm{s}$ sum of $\triangle A C D$ )
$\therefore \triangle A C D$ is a right-angled isosceles triangle.
$C D=A D=n$
(sides opp. equal angle)
$E G \perp C D \Rightarrow E G / / A D \Rightarrow \triangle C E G \sim \triangle C A D$ (equiangular)
$\Rightarrow C G=E G=4$
(ratio of sides, $\sim \Delta$ 's)
$$
n=A D=C D=3+3+4=10
$$

Method 2
Draw $E G / / C D$, which intersects $A B$ at $G$.
$G D=B D=4 \quad(B F=F E$ and $F D / / E G$, intercept theorem $)$
$G E=2 D F=6 \quad$ (mid-points theorem)
$\angle A G E=90^{\circ} \quad$ (corr. $\left.\angle \mathrm{s}, E G / / C D\right)$
$\triangle A G E$ is a right-angled isosceles triangle.
$\therefore A G=G E=6$ (sides opp. eq. $\angle$ s)
$$
n=A G+G D=6+4=10
$$