3. Let $\Varangle A O B, \Varangle B O C, \Varangle C O D, \Varangle D O E, \Varangle E O A$ be angles around point $O$ such that: $m(\Varangle B O C)=m(\Varangle A O B)+n^{0}, m(\Varangle C O D)=m(\Varangle B O C)+n^{0}, m(\Varangle D O E)=m(\Varangle C O D)+n^{0}$, $m(\nless E O A)=m(\nless D O E)+n^{0}, n \in \mathbb{N}^{*}$, and $(O A$ and $(O D$ are opposite rays.

a) (3p) determine $m(\Varangle A O B)$;

b) (4p) if $(OM$ is the bisector of $\Varangle A O B$ and $(O N$ is the bisector of $\Varangle C O D$, and $[O M] \equiv[O E]$, show that $[M N] \equiv[N E]$.

Prof. Stela Boghian
Solution: a) Let $m(\angle A O B)=x, m(\angle A O D)=m(\angle D O A)$, $x+x+n^{0}+x+2 n^{0}=x+3 n^{0}+x+4 n^{0}$, from which $x=4 n^{0}$, $m(\angle A O B)=4 n^{0} ; m(\angle B O C)=5 n^{0} ; m(\angle C O D)=6 n^{0}$, together they have $180^{\circ} \Rightarrow n^{0}=12^{\circ}$ and $m(\angle A O B)=48^{\circ}$.
b) $m(\angle B O C)=60^{\circ} ; m(\angle C O D)=72^{\circ} ; m(\angle D O E)=84^{\circ}$; $m(\angle E O A)=96^{\circ} \Rightarrow m(\angle M O N)=m(\angle E O N)=120^{\circ}$. $\triangle M O N \equiv \triangle E O N($ L.U.L. $) \Rightarrow[M N] \equiv[N E]$.

![](https://cdn.mathpix.com/cropped/2024_06_07_7df4e3f3ae1f532ed64fg-2.jpg?height=357&width=491&top_left_y=1786&top_left_x=1362)

## Grading Rubric:

| a) Let $m(\angle A O B)=x, m(\angle A O D)=m(\angle D O A), x+x+n^{0}+x+2 n^{0}=x+3 n^{0}+x+4 n^{0}$ | 2p |
| :--- | :--- |
| $n^{0}=12^{0}$ and $m(\angle A O B)=48^{0}$. | $\mathbf{1 p}$ |
| b) $m(\angle B O C)=60^{\circ} ; m(\angle C O D)=72^{\circ} ; m(\angle D O E)=84^{\circ} ; m(\angle E O A)=96^{\circ}$ | $\mathbf{1 p}$ |
| $m(\angle M O N)=m(\angle E O N)=120^{\circ}$. | $\mathbf{1 p}$ |
| Prove that $\triangle M O N \equiv \triangle E O N$ | $\mathbf{1 p}$ |
| $[M N] \equiv[N E]$ | $\mathbf{1 p}$ |