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 $E_1{F}_1\parallel AC\parallel H_1{G}_1$. That means $\frac{F_{1}C}{C{G}_1}=\frac{E_{1}A}{A{H}_1}=\lambda$.

(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 $E_1{H}_1$ at $M$ and $N$ respectively. As $E{B}_1\parallel EF$, we get $E_{1}A=\frac{BA}{EA}\cdot AM$. In the same way, we get $H_{1}A=\frac{AD}{AH}\cdot AN$. Then $$\frac{E_{1}A}{H_{1}A}=\frac{AM}{AN}\cdot \frac{AB}{AE}\cdot \frac{AH}{AD}.\stackrel{◯}{1}$$ $$\text{On the other hand, }\frac{AM}{AN}=\frac{EQ}{QH}=\frac{△AEC}{△AHC}=\frac{△ABC}{△ADC}\cdot \frac{AE}{AB}\cdot \frac{AD}{AH}.\stackrel{◯}{2}$$ From ①, ② we get $\frac{E_{1}A}{H_{1}A}=\frac{EQ}{QH}\cdot \frac{AB}{AE}\cdot \frac{AH}{AD}=\frac{△ABC}{△ADC}$. In the same way, $$\frac{F_{1}C}{C{G}_1}=\frac{△ABC}{△ADC}.$$ $$\text{So }\frac{F_{1}C}{C{G}_1}=\frac{E_{1}A}{A{H}_1}=\lambda .$$