Trouble understanding a geometry problem involving triangle and bisector.

Solution 1:

As @dxiv pointed out in the comment, the statement means $C$ is midpoint of $AB$. In that case part i) becomes trivial.

For part ii) when $OP$ does not pass through $C$ we can use similarity to prove the statement. Note that the question is asking the "difference of $AX$ and $BY$", this means absolute value of the difference as you can see below.

Suppose, $OP$ passes through $AC$ at $M$. We have triangles $AMX$, $CMZ$, and $BMY$ all similar to each other. By similarity $$\frac{AX}{CZ}=\frac{AC-MC}{MC}, \frac{BY}{CZ}=\frac{BC+MC}{MC}$$

Since $AC=BC$, subtracting the two we get the result.