4. 

![](https://cdn.mathpix.com/cropped/2024_06_07_9449e85eb4ec58d4f57cg-2.jpg?height=968&width=1151&top_left_y=1669&top_left_x=567)
a) If $[O Q$ is the bisector of $\widehat{A O M}$, and $[O S$ is the bisector of $\widehat{B O N}$, then $m(\widehat{A O Q}) + m(\widehat{B O S}) = 120^{\circ} - 90^{\circ} = 30^{\circ}$, so $m(\widehat{A O M}) + m(\widehat{B O N}) = 60^{\circ}$ and $m(\widehat{M O N}) = 120^{\circ} - 60^{\circ} = 60^{\circ}$ $4 p$

b) It will result that $m(\widehat{N O P}) = 180^{\circ} - m(\widehat{M O N}) = 120^{\circ}$ and $\widehat{N O P} \equiv \widehat{A O B}$, which means $\widehat{B O P} + \widehat{B O N} = \widehat{A O N} + \widehat{B O N}$, so $\widehat{B O P} \equiv \widehat{A O N}$ $3 p$
4. Let $\overline{A O B}$ be an angle with a measure of $120^{\circ}$, and let $M$ and $N$ be two points inside it such that $M$ is inside the angle $\overline{A O N}$, and the bisectors of the angles $\overline{A O M}$ and $\overline{B O N}$ are perpendicular.

a) Determine the measure of the angle $\overline{M O N}$.

b) If $P$ is a point such that $O \in(M P)$, show that $\widehat{B O P} \equiv \widehat{A O N}$

## Grading Rubric for Grade 6

