3. Consider the rays [OA, [OB, [OC, [OD, and [OE in this order. If $m(\angle B O C)+10^{0}=m(\angle A O B) ; 3 \cdot m(\angle C O D)=4 \cdot m(\angle B O C) ; m(\angle A O E)=$ $2 m(\angle D O E) ; 3 \cdot m(\angle A O E)=5 \cdot m(\angle B O C)$, then:

a) Find $m(\angle A O B), m(\angle B O C), m(\angle C O D), m(\angle D O E)$, and $m(\angle E O A)$;

b) If [OF is the bisector of $\angle A O E$, show that points F, O, and C are collinear.
3.a) $\frac{m(\angle COD)}{4}=\frac{m(\angle BOC)}{3}=k \rightarrow m(\angle BOC)=3k \text{ and } m(\angle COD)=4k \rightarrow m(\angle AOB)=3k+10^{\circ}$. From $3 \cdot m(\angle AOE)=5 \cdot m(\angle BOC)$ we have that $m(\angle AOE)=5k$ and $m(\angle DOE)=\frac{5k}{2}$, then from their sum we have that $k=20$ so $m(\angle AOB)=70^{\circ}, m(\angle BOC)=60^{\circ}, m(\angle COD)=80^{\circ}, m(\angle DOE)=50^{\circ}, m(\angle AOE)=100^{\circ}$.

5 points.

b) $[OF]$ bisector. Then $m(\angle AOF)=50^{\circ}, m(\angle AOC)=70^{\circ}+60^{\circ}=130^{\circ}$ and $m(\angle FOC)=180^{\circ}$. Then F, O, C are collinear.

2 points.