[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.