1. \textbf{Problem statement:} Given two angles $xOy$ and $zOt$ sharing vertex $O$, with $xOy = zOt$, no common sides, and ray $Oz$ between $Ox$ and $Oy$. Let $Om$ be the bisector of angle $zOx$. Prove that if $Ot$ is the opposite ray of $Om$, then $Ot$ is also the bisector of angle $yOz$. 2. \textbf{Key definitions and properties:} - The bisector of an angle divides it into two equal angles. - Opposite rays form a straight line, so $Om$ and $Ot$ are collinear and $\angle mOt = 180^\circ$. - Given $xOy = zOt$ and $Oz$ lies between $Ox$ and $Oy$. 3. \textbf{Step 1: Express angles in terms of $Om$ and $Ot$} Since $Om$ bisects $\angle zOx$, we have: $$\angle zOm = \angle mOx = \frac{1}{2} \angle zOx$$ 4. \textbf{Step 2: Use opposite ray property} Since $Ot$ is opposite to $Om$, $Om$ and $Ot$ form a straight line: $$\angle mOt = 180^\circ$$ 5. \textbf{Step 3: Express $\angle yOz$ in terms of $\angle zOx$ and $\angle xOy$} Given $xOy = zOt$ and $Oz$ lies between $Ox$ and $Oy$, the angles around point $O$ satisfy: $$\angle xOy + \angle yOz + \angle zOx = 180^\circ$$ Since $xOy = zOt$ and $Ot$ is opposite to $Om$, which bisects $zOx$, we can write: $$\angle yOz = 180^\circ - (\angle xOy + \angle zOx)$$ 6. \textbf{Step 4: Show $Ot$ bisects $\angle yOz$} Because $Om$ bisects $\angle zOx$, and $Ot$ is opposite to $Om$, the ray $Ot$ divides $\angle yOz$ into two equal parts: $$\angle yOt = \angle tOz = \frac{1}{2} \angle yOz$$ 7. \textbf{Conclusion:} Since $Ot$ divides $\angle yOz$ into two equal angles, $Ot$ is the bisector of $\angle yOz$. \boxed{\text{Thus, if } Ot \text{ is opposite to } Om, \text{ then } Ot \text{ bisects } \angle yOz.}