在下图中，点 A、B 和 C 位于圆上。AC 是以 O 为圆心的圆的直径。 圆在 X 点的切线与 AC 的延长线交于 Y 点。 弦 XB 与 AC 相交于 F 点。 XC 平分角 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:
$AC \times FC = BC \times BC$
$AC \times FC = BC^{2}$