No one solved this problem completely although Andrei Lazanu's solution came close. It is not necessary to redraw the diagram to complete the set of inequalities. Perhaps you could use the same argument for the last part Andrei.

**Part of Andrei's solution**

I first observed that a/b is the tangent of angle BAC in Figure 1 and c/d the tangent of angle FDB.

I also see that the measure of angle BAC is greater than the measure of angle DAE.

All angles of interest for the problem are smaller than 90 degrees, and in this region the tangent is a monotonic increasing function (it does not decrease in this range), so the order relation between angles is transferred to tangents.

Now I write the tangents of these angles: $${tan(DAE)} = {DE \over AE} = {a \over b}$$ $${tan(BAC)} = {BC \over AC} = {{a+c} \over {b+d}}$$I have now to transpose the relation between the angles to their tangents, and I obtain the inequality $${a \over b} < { {a+c} \over {b+d}}$$ for $${a \over b} < {c \over d}$$