Problem 1790

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,
$$\begin{aligned} AP^2 &= AQ \times AP + PQ \times AP \\ &= AF \times AD + PF \times PE \\ &= (AO^2 - r^2) + (PO^2 - r^2). \end{aligned}$$
①

Similarly,
$$AG^2 = (AO^2 - r^2) + (GO^2 - r^2).$$
②

By ①, ②, we have $AP^2 - AG^2 = PO^2 - GO^2$, which implies that $PG \perp AO$.

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

Fig. 2

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

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

③ $\div$ ④ yields
$$\frac{DA}{AF} = \frac{DK}{KF}. \quad \textcircled{5}$$

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
$$AK \times FD = AF \times KD + AD \times FK = 2AF \times KD. \quad \textcircled{6}$$

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,
$$\frac{CP}{PD} = \frac{AF}{FD}. \quad \textcircled{7}$$

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

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

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