China Mathematical Olympiad

(1) If $EF \parallel AC$, then $\frac{BE}{EA} = \frac{BF}{FC}$.
So $\frac{DH}{HA} = \frac{DG}{GC}$ using condition (a). Then $HG \parallel AC$, giving $E_1F_1 \parallel AC \parallel H_1G_1$. That

means $\frac{F_1C}{CG_1} = \frac{E_1A}{AH_1} = \lambda$.

(2) If $EF$ is not parallel to $AC$, extend lines $EF$ and $AC$ to meet at point $T$. By Menelaus' Theorem, we have $\frac{CF}{FB} \cdot \frac{BE}{EA} \cdot \frac{AT}{TC} = 1$, and then $\frac{CG}{GD} \cdot \frac{DH}{HA} \cdot \frac{AT}{TC} = 1$ using condition (a). By the inverse of Menelaus' Theorem, we know that points $T$, $H$ and $G$ are collinear. Suppose lines $TF$ and $TG$ meet line $E_1H_1$ at $M$ and $N$ respectively. As $EB_1 \parallel EF$, we get $E_1A = \frac{BA}{EA} \cdot AM$. In the same way, we get $H_1A = \frac{AD}{AH} \cdot AN$. Then

$$\frac{E_1A}{H_1A} = \frac{AM}{AN} \cdot \frac{AB}{AE} \cdot \frac{AH}{AD}. \qquad \textcircled{1}$$
$$\text{On the other hand, } \frac{AM}{AN} = \frac{EQ}{QH} = \frac{\triangle AEC}{\triangle AHC} = \frac{\triangle ABC}{\triangle ADC} \cdot \frac{AE}{AB} \cdot \frac{AD}{AH}. \qquad \textcircled{2}$$
From ①, ② we get $\frac{E_1A}{H_1A} = \frac{EQ}{QH} \cdot \frac{AB}{AE} \cdot \frac{AH}{AD} = \frac{\triangle ABC}{\triangle ADC}$. In the same way,
$$\frac{F_1C}{CG_1} = \frac{\triangle ABC}{\triangle ADC}.$$
$$\text{So } \frac{F_1C}{CG_1} = \frac{E_1A}{AH_1} = \lambda.$$