Question: Mối quan hệ các đường thẳng trong mặt phẳng 3) Given $\angle xOy = 60^\circ$, $A \in Ox$. Through $A$, draw $Ay' \parallel Oy$ and lie inside $\angle xOy$. $Ot$ and $At'$ are bisectors of $\angle xOy$ and $\angle xAy'$ respectively. Cho $xOy = 60^\circ$, điểm $A\in Ox$, qua $A$ dựng $Ay' \parallel Oy$ và nằm trong $xOy$. Gọi $Ot$ và $At'$ là phân giác của $xOy$ và $xAy'$. a) Find $\angle xAy'$ and $\angle OAy'$. b) Find $\angle xAt'$ and $\angle AOt$. c) Prove that: $Ot \parallel At'$.

1. Problem: Given $\angle xOy = 60^\circ$, $A \in Ox$, and $Ay' \parallel Oy$ inside $\angle xOy$. $Ot$ and $At'$ are bisectors of $\angle xOy$ and $\angle xAy'$ respectively. 2. Find $\angle xAy'$ and $\angle OAy'$: - Since $Ay' \parallel Oy$ and $A \in Ox$, $\angle xOy = 60^\circ$. - $\angle xAy' = \angle xOy = 60^\circ$ (alternate interior angles). - $\angle OAy' = 180^\circ - \angle xAy' = 180^\circ - 60^\circ = 120^\circ$. 3. Find $\angle xAt'$ and $\angle AOt$: - $Ot$ bisects $\angle xOy = 60^\circ$, so $\angle xOt = \angle tOy = \frac{60^\circ}{2} = 30^\circ$. - $At'$ bisects $\angle xAy' = 60^\circ$, so $\angle xAt' = \angle t'Ay' = \frac{60^\circ}{2} = 30^\circ$. - $\angle AOt = \angle xOt = 30^\circ$. 4. Prove $Ot \parallel At'$: - $Ot$ and $At'$ both form $30^\circ$ angles with $Ox$. - Since $Ot$ and $At'$ are both at $30^\circ$ to $Ox$, they are parallel. Final answers: - $\angle xAy' = 60^\circ$, $\angle OAy' = 120^\circ$. - $\angle xAt' = 30^\circ$, $\angle AOt = 30^\circ$. - $Ot \parallel At'$ proven by equal angles with $Ox$.