In $\bigtriangleup ABC$, the points $A_2, A_3, A_4$ are chosen inside the triangle so that $\angle ABA_2 = \angle A_2BA_3 = \angle A_3BA_4 = \angle A_4BC.$ If $A, A_2, A_3, A_4$ are collinear, find $\angle BA_2A_3$ given that $\angle A = 50.$
$A. 85/2$
$B. 40$
$C. 75/2$
D. Not uniquely determined
E. None of given
Here's my diagram (letting $A, A_2, A_3, A_4$ be on $AC$):
So I first let $\angle ABA_2=x^{\circ}$. Then, $\angle C=130-4x^{\circ}$. Then, $\angle ABA_{i}=130-(i-1)x^{\circ}$. So, $\angle BA_2A_3=50+x^{\circ}$. That narrows it down to D or E, and I guessed D, since it looked like $x$ had no restrictions. However, the answer was E.
Please post a solution. Thanks.