I4.4 In Figure $2, E F G$ is a right-angled triangle. Given that $H$ is a point on $F G$, such that $G H: H F=4: 5$ and $\angle G E H=\angle F E H$. If $E G=c$ and $F G=d$, find the value of $d$.
Let $\angle F E H=\theta=\angle F E H, G H=4 k, F H=5 k, E G=1$
In $\triangle E G H, \tan \theta=4 k \ldots \ldots$ (1)
In $\triangle E F G, \tan 2 \theta=9 k$. $\qquad$
Sub. (1) into (2): $\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta} \Leftrightarrow 9 k=\frac{2 \cdot 4 k}{1-(4 k)^{2}}$
$9\left(1-16 k^{2}\right)=8 \Leftrightarrow k=\frac{1}{12} \Leftrightarrow d=F G=5 k+4 k=9 k=\frac{3}{4}$
Method 2 Let $\angle G E H=\theta=\angle F E H$
From $P$, draw a line segment $F P$ parallel to $G E$, which intersects with $E H$ produced at $P$.
$\angle F P H=\theta$
(alt. $\angle \mathrm{s}, P F / / G E$ )
$\triangle F P H \sim \triangle G E H$
(equiangular)
$$
\begin{array}{l}
\frac{G H}{H F}=\frac{G E}{P F} \Rightarrow \frac{4}{5}=\frac{1}{P F} \Rightarrow P F=1.25 \\
F E=P F=1.25 \\
d=F G=\sqrt{1.25^{2}-1}=0.75
\end{array}
$$
(ratio of sides, $\sim \Delta$ 's)
(sides opp. eq. $\angle$ s)
(Pythagoras' Theorem)