Example 5 As shown in Figure 5, given $\triangle A B C$ with two altitudes $A D$ and $B E$ intersecting at point $H$, the circumcenter of the triangle is $O$. A line $O F \perp B C$ is drawn through $O$ and intersects $B C$ at point $F$, and $O H$ intersects $A F$ at point $G$. Then $S_{\triangle O F G}: S_{\triangle C H A}=(\quad)$.
(A) $1: 4$
(B) $1: 3$
(C) $2: 5$
(D) $1: 2$
(2010, Joint Autonomous Admission Examination of Tsinghua University and Other Universities)
Since $O$ is the circumcenter of $\triangle A B C$ and $O F \perp B C$, therefore, $F$ is the midpoint of side $B C$.
Thus, $A F$ is the median on side $B C$.
By Euler's theorem, the intersection point $G$ of $O H$ and $A F$ is the centroid of $\triangle A B C$.
Therefore, $F G: G A=1: 2$.
Also, $\triangle O F G \backsim \triangle H A G$, so the ratio of the areas of the two triangles is $1: 4$.