Consider the triangles <B0A1> and <A0B1>. Angle B0A1 and A0B1 are the same (vertically opposite) and we know AB1 in terms of B1C and BA1 in terms of A1C.
The sine rule relates BOA1 angle to line BA1 with angle A1B0 to line A1O and similarly angle AOB1 to line AB1 and angle AB1O to line AO.
Angles BOA1 and AOB1 are the same so we can write the relationship as:
sin(BA01)/BA1 = sin(A1B0)/A10 and sin(A0B1)/AB1 = sin(AB10)/AO but sin(BA01) = sin(A0B1) which implies
BA1*sin(A1B0)/A10 = sin(AB10)*AB1/AO which means
AO/A10 = sin(AB10)/sin(A1B0) * [AB1/BA1]
We are given BA1 = A1C/p and AB1 = B1C/q so AB1/BA1 = p/q * (B1C/A1C)
The angles AB10 and A1B0 are the same and we can discuss this if you want (it involves doing supplementary angle proofs along the lines AC and BC).
So we have:
A0/A10 = AB1/BA1 = p/q * (B1C/A1C)
Another application of the sine rule and the fact that sin(180-x) = sin(x) (in degrees) gives B1C = A1C which finally gives the result:
A1/A10 = p/q.
You probably have learned in class a set of rules to do this that are a lot easier but I haven't done this kind of thing for more than 15 years so I'm sorry if this is long winded.