In the figure below, points A, B and C lie on the circle. AC is a diameter of a circle with centre O. The tangent to the circle at X meets AC produced at Y. Chord XB intersects AC at F. XC bisects the angle YXB.

1. Identify the relationship between the angle formed by the tangent and chord, and the angle in the alternate segment.

Let $\angle YXC = \theta$.
By the alternate segment theorem, the angle between the tangent $XY$ and the chord $XC$ is equal to the angle in the alternate segment.
Therefore, $\angle XBC = \angle YXC = \theta$.

2. Use the angle bisector property to find another angle equal to theta.

Since $XC$ bisects $\angle YXB$, we have $\angle CXB = \angle YXC = \theta$.

3. Conclude that triangle CXB is isosceles, hence CX = CB.

In $\triangle CXB$, we have $\angle XBC = \theta$ and $\angle CXB = \theta$.
Since $\angle XBC = \angle CXB$, $\triangle CXB$ is an isosceles triangle.
Therefore, $CX = CB$.

1. Show that the first pair of angles are equal using angles in the same segment.

$\angle XAC = \angle XBC$ (angles in the same segment).
Since $F$ lies on the chord $XB$, $\angle XBC = \angle FBC$.
Therefore, $\angle XAC = \angle FBC$.

2. Express $\angle FCB$ in terms of $\angle CXB$.

$\angle AXC = 90^\circ$ (angle in a semicircle).
$\angle AXB = \angle AXC - \angle CXB = 90^\circ - \angle CXB$.
$\angle FCB = \angle ACB = \angle AXB$ (angles in the same segment).
So, $\angle FCB = 90^\circ - \angle CXB$.

3. Express $\angle XCA$ in terms of $\angle CXB$ and conclude similarity.

By the alternate segment theorem, $\angle YXC = \angle XAC$.
Since $XC$ bisects $\angle YXB$, $\angle YXC = \angle CXB$, which means $\angle XAC = \angle CXB$.
In $\triangle AXC$, $\angle XCA = 180^\circ - 90^\circ - \angle XAC = 90^\circ - \angle CXB$.
Therefore, $\angle XCA = \angle FCB$.
Since $\angle XAC = \angle FBC$ and $\angle XCA = \angle FCB$, $\triangle AXC$ is similar to $\triangle BFC$ (AA similarity).

1. Set up the ratio of corresponding sides using the similar triangles from part (b)

Since $\triangle AXC$ is similar to $\triangle BFC$, the ratio of their corresponding sides is equal:
$\frac{AC}{BC} = \frac{XC}{FC}$

2. Substitute the result from part (a) to complete the proof

Cross-multiplying gives:
$AC \times FC = BC \times XC$
From part (a), we know that $XC = BC$. Substituting this into the equation:
$AC \times FC = BC \times BC$
$AC \times FC = BC^{2}$