China National Team Selection Test

As shown in Fig. 1, extend $EF$ and meet $BC$ at point $P$, and join and extend $GP$, which meets $AD$ at $K$ and the extension of $AC$ at $L$.

Fig. 1

As shown in Fig. 2, let $Q$ be a point on $AP$ such that $$\angle PQF=\angle AEF=\angle ADB.$$ It is easy to see that $A,E,F,Q$ and $F,D,P,Q$ are concyclic respectively. Denote by $r$ the radius of $\odot O$. By the power of a point theorem, $$\begin{array}{rl}A{P}^2& =AQ\times AP+PQ\times AP\\ & =AF\times AD+PF\times PE\\ & =(A{O}^2-r^2)+(P{O}^2-r^2).\end{array}$$ ①

Similarly, $$A{G}^2=(A{O}^2-r^2)+(G{O}^2-r^2).$$ ②

By ①, ②, we have $A{P}^2-A{G}^2=P{O}^2-G{O}^2$, which implies that $PG\perp AO$.

As shown in Fig. 3, applying Menelaus' theorem for $△PFD$ and line $AEB$, we have

Fig. 2

$$\frac{DA}{AF}\times \frac{FE}{EP}\times \frac{PB}{BD}=1.\stackrel{◯}{3}$$

Applying Ceva's theorem for $△PFD$ and point $G$, we have $$\frac{DK}{KF}\times \frac{FE}{EP}\times \frac{PB}{BD}=1.\stackrel{◯}{4}$$

③ $÷$ ④ yields $$\frac{DA}{AF}=\frac{DK}{KF}.\stackrel{◯}{5}$$

Fig. 3

Equation ⑤ illustrates that $A,K$ are harmonic points with respect to $F,D$, i.e. $AF\times KD=AD\times FK$.

It follows that $$AK\times FD=AF\times KD+AD\times FK=2AF\times KD.\stackrel{◯}{6}$$

Since $B,D,F,E$ are concyclic, we have $\angle DBA=\angle EFA$. Since $\angle CAD=\angle CBA$, we have $\angle CAF=\angle EFA$, which implies that $AC\parallel EP$. Thus, $$\frac{CP}{PD}=\frac{AF}{FD}.\stackrel{◯}{7}$$

Applying Menelaus' theorem to $△ACD$ and line $LPK$, we have $$\frac{AL}{LC}\times \frac{CP}{PD}\times \frac{DK}{KA}=1.\stackrel{◯}{8}$$

Combining ⑥, ⑦ and ⑧, we have $\frac{AL}{LC}=2$.

In $△AGL$, $M,C$ are the midpoints of $AG,AL$ respectively, and hence $MC\parallel GL$. Since $GL\perp AO$, we conclude that $MC\perp AO$.