Shayan-TayefehIR wrote: In the figure below points $A,B$ are the centers of the circles $\omega_1, \omega_2$. Starting from the line $BC$ points $E,F,G,H,I$ are obtained respectively. Find the angle $\angle IBE$. Please add the picture for this problem:

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Our general strategy will be to keep track of \( \angle APB \) for points \( P \) mentioned in the problem. While this problem is not particularly difficult, this approach provides a relatively systematic solution. We begin with two lemmas. Lemma 1: If \( P \in w_1 \), \( Q \in w_2 \), and $C$ lies on segment $PQ$, then \[ \angle APB + \angle AQB = 60^\circ. \]

Proof

Lemma 2: If \( P \in w_1 \), \( Q \in w_2 \), and \( P \), \( A \), and \( Q \) are collinear, then \[ \angle APB = \frac{1}{2} \angle AQB.

\]Proof

Using these two lemmas (and their variants), the problem proceeds smoothly: \[ \angle ACB = 60^\circ \implies \angle AEB = 30^\circ \implies \angle AFB = 15^\circ \implies \angle AGB = 45^\circ \implies \angle AHB = 22.5^\circ \implies \angle AIB = 37.5^\circ. \] Finally, to complete the problem: \[ \angle IBE = 2\angle IAE = 2(\angle AIB - \angle AEB) = 2(37.5^\circ - 30^\circ) = \boxed{15^\circ}. \]