Draw the triangle, and then draw it again rotated so that B is in the same corner that C was. Now since ICA < IBA and ICB < IBC, ICA + ICB < IBA + IBC. Since this is true, the angle at C is less than the angle at B. I forget the name of the proof, but if you have such a triangle, the side with a greater angle has a smaller diagonal (BB' and CC'). So Since angle C < B, CC' > BB'.

I don't understand quite understand the wording in 2.