[Case 3 of Lemma 11.6.1, p.190]

3. If α is not acute but |ac| > |ab|, roll I along ac unti c meets a. Thus ac is creased at its midpoint a'. Let c' be the other end of the crease, where the perpendicular bisector of ac meets bc. Now [triangle]a'c'b is an ear to which Case 1 applies, rolling that ear into P. The new ear is [triangle]aba', with the same α but with |aa'| half the length of |ac|. Repeat until Case 2 applies.