In the figure, $BC$ is a diameter of the circle with $O$ as the centre. $H$ is the midpoint of $CF$.
$ABC$ is a straight line and $AG$ is a tangent to the circle at point $F$. The line $EF$ intersects
$OH$ at point $X$ and $E$ is the midpoint of $CO$.

1. Identify a shared angle between the two triangles

In $\triangle ABF$ and $\triangle AFC$, $\angle FAB = \angle FAC$ (common angle).

2. Use the alternate segment theorem to find a second pair of equal angles

Since $AG$ is a tangent to the circle at $F$ and $FC$ is a chord, by the alternate segment theorem, $\angle AFB = \angle ACF$.

3. Conclude similarity using the AA criterion

Since two pairs of corresponding angles are equal, $\triangle ABF$ is similar to $\triangle AFC$ (AA similarity).

1. Use the similarity of triangles from part (i) to set up a ratio of corresponding sides.

Since $\triangle ABF$ is similar to $\triangle AFC$, the ratio of their corresponding sides is equal:
$\frac{AB}{AF} = \frac{AF}{AC}$

2. Express $AF^2$ in terms of $AB$ and $AC$, then substitute $AC$ using segment addition.

Cross-multiplying gives:
$AF^2 = AB \times AC$
Since $A, B,$ and $C$ lie on a straight line in that order, $AC = AB + BC$. Substituting this into the equation:
$AF^2 = AB(AB + BC)$
$AF^2 = AB^2 + AB \times BC$

1. Apply the Midpoint Theorem to triangle COF.

In $\triangle COF$, $E$ is the midpoint of $CO$ and $H$ is the midpoint of $CF$.
By the Midpoint Theorem, $EH \parallel OF$ and $EH = \frac{1}{2}OF$.

2. Identify equal angles to prove similarity between triangles OXF and HXE.

Consider $\triangle OXF$ and $\triangle HXE$:
$\angle OXF = \angle HXE$ (vertically opposite angles)
$\angle XOF = \angle XHE$ (alternate angles, $EH \parallel OF$)
$\angle XFO = \angle XEH$ (alternate angles, $EH \parallel OF$)

3. State the similarity of the triangles.

Therefore, $\triangle OXF$ is similar to $\triangle HXE$.

4. Use the ratio of corresponding sides to find the ratio of OX to XH.

Since the triangles are similar, the ratio of their corresponding sides is equal:
$\frac{OX}{XH} = \frac{OF}{EH}$
Since $EH = \frac{1}{2}OF$, we have $\frac{OF}{EH} = \frac{2}{1}$.
Thus, $\frac{OX}{XH} = \frac{2}{1}$, which means $OX : XH = 2 : 1$.