## SUBJECT III

Given the collinear points A, $\mathrm{O}, \mathrm{D}$ where $\mathrm{O} \in(\mathrm{AD})$ and the adjacent angles $\Varangle \mathrm{AOB}$ and $\Varangle \mathrm{BOC}$, with the ray (OC) being interior to the angle $\Varangle \mathrm{BOD}$. If $\mathrm{m}(\Varangle \mathrm{BOC})=5 \cdot \mathrm{m}(\Varangle \mathrm{AOB})$, $\mathrm{m}(\Varangle \mathrm{BOC})=\frac{5}{3} \mathrm{~m}(\Varangle \mathrm{COD})$ and $[O M$ is the bisector of $\Varangle \mathrm{AOC}$, and $\mathrm{Q}$ is a point interior to the angle $\Varangle \mathrm{BOD}$ such that $\mathrm{m}(\Varangle \mathrm{MOQ})=90^{\circ}$, the following is required:
a) $\mathrm{m}(\Varangle \mathrm{AOB}), \mathrm{m}(\Varangle \mathrm{BOC}), \mathrm{m}(\Varangle \mathrm{COD})$.

b) Show that [OQ is the bisector of the angle m( $\Varangle \mathrm{COD}$ ).
## Solution:

a) Let $\mathrm{m}(\angle \mathrm{BOC})=\mathrm{a}$, so $\mathrm{m}(\angle \mathrm{AOB})=\frac{a}{5}, \mathrm{~m}(\angle \mathrm{COD})=\frac{3 a}{5}$

$$
\begin{gathered}
\text { We obtain the relation } \frac{a}{5}+a+\frac{3 a}{5}=180^{\circ} \\
\mathrm{a}=100^{\circ}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_00e82414df2d953fbe02g-4.jpg?height=283&width=305&top_left_y=121&top_left_x=356)
Conclusion $\mathrm{m}(\angle \mathrm{AOB})=20^{\circ}, \mathrm{M}(\angle \mathrm{BOC})=100^{\circ}, \mathrm{m}(\angle \mathrm{COD})=60^{\circ}$ ..... $2 p$
b) $\mathrm{m}(\angle \mathrm{AOM})=\mathrm{m}(\angle \mathrm{AOC}): 2=60^{\circ}$ ..... $1 p$
$\mathrm{m}(\angle \mathrm{QOD})=180^{\circ}-[\mathrm{m}(\angle \mathrm{AOM})+\mathrm{m}(\angle \mathrm{MOQ})]=$
$=180^{\circ}-\left(60^{\circ}+90^{\circ}\right)=30^{\circ}$ ..... $1 p$

$\mathrm{m}(\angle \mathrm{QOD})=30^{\circ}=60^{\circ}: 2=\mathrm{m}(\angle \mathrm{COD}): 2$, so (OQ is the bisector of $\angle \mathrm{COD}$