# No, its impossible

Geometry Level 4

Suppose I have $$\angle ABC$$, and I draw out an angle bisector of $$\angle ABC$$, such that $$\angle ABA_{1} = \angle CBA_{1}$$ as described above.

Next, I make an angle bisector of $$\angle CBA_{1}$$, such that $$\angle A_{1}BA_{2} = \angle A_{2} BC$$

Next, I make an angle bisector of $$\angle A_{1}BA_{2}$$, such that $$\angle A_{1}BA_{3} = \angle A_{3}BA_{2}$$

Next, I make an angle bisector of $$\angle A_{3}BA_{2}$$, such that $$\angle A_{3}BA_{4} = \angle A_{4}BA_{2}$$

If I am to repeat this algorithm forever, find $\lim _{ n\rightarrow \infty }{ \frac { \angle { A }_{ n }BC }{ \angle { A }BC } }$