I believe I've spent way to much time trying to grasp a certain part of this proof. This is provided in S.L. Loney's book on The Elements of Coordinate Geometry.
Let LPL' be a straight line which cuts the axis of Y at a distance c from the origin and is inclined at an angle θ to the axis of X.
Let P be any point on the straight line. Draw PNM parallel to the axis of y to meet OX in M, and let it meet the straight line through C parallel to the axis of x in the point N.
Let P be the point (x,y), so that CN = 0M = x, and NP = MP – OC = y – c.
Since CPN = PNN' – PCN' = ω – θ, we have $y – c \over x$$ = $$NP \over CN $$ = $$sinNCP \over sinCPN$$ = $$sinθ \over sin(ω – θ)$.
I'll stop here, for this is the part I can't at all grasp. Why exactly is $NP \over CN $$ = $$sinNCP \over sinCPN$. For we are currently talking of oblique coordinates, and there is no right angle to orient ourselves with. Perhaps this question is laughable; but I would still appreciate the reader if I were properly answered by them. Below is provided a picture for visual aid: 
Thank you in advance.