I4.4 In Figure 1 , let $O D C$ be a triangle. Given that $F H, A B$, $A C$ and $B D$ are line segments such that $A B$ intersects $F H$ at $G, A C, B D$ and $F H$ intersect at $E, G E=1, E H=c$ and $F H / / O C$. If $d=E F$, find the value of $d$.
Remark: there are some typing mistakes in the Chinese old version: ... AC 及 $A D$ 為綫段 ...FH// $B C \ldots$
$\triangle A G E \sim \triangle A B C$ (equiangular)
Let $\frac{C E}{A E}=k, A E=x, A G=t$.
$B C=k+1, E C=k x, G B=k t$ (ratio of sides, $\sim \Delta$ 's)
$\triangle D E H \sim \triangle D B C$ (equiangular)
$\frac{B C}{E H}=\frac{k+1}{2}=\frac{D B}{D E} \quad$ (ratio of sides, $\sim \Delta$ 's)
Let $D E=2 y \Rightarrow D B=(k+1) y$
$E B=D B-D E=(k-1) y$
$\triangle A F G \sim \triangle A O B$ (equiangular)
$F G=d-1, \frac{O B}{F G}=\frac{A B}{A G}$ (ratio of sides, $\sim \Delta$ 's)
$O B=(d-1) \cdot \frac{(k+1) t}{t}=(d-1)(k+1)$
$\triangle D F E \sim \triangle D O B$ (equiangular)
$\frac{F E}{O B}=\frac{D E}{D B}$ (ratio of sides, $\sim \Delta$ 's)
$\Rightarrow d=(d-1)(k+1) \cdot \frac{2 y}{(k+1) y} \Rightarrow d=2$
Method 2
$\triangle A F G \sim \triangle A O B$ and $\triangle A G E \sim \triangle A B C$
$\frac{d-1}{1}=\frac{O B}{B C}$ (ratio of sides, $\sim \Delta$ 's)
$\triangle D F E \sim \triangle D O B$ and $\triangle D E H \sim \triangle D B C$
$\frac{d}{2}=\frac{O B}{B C}$ (ratio of sides, $\sim \Delta$ 's)
Equating the two equations
$\frac{d-1}{1}=\frac{d}{2}$
$d=2$