## Subject III

Lines $A B$ and $C D$ intersect at point $O$, and m( $\angle \mathrm{AOD})<90^{\circ}$.

Let [OM, [ON and [OP be the interior bisectors of angles $\angle A O D, \angle M O B$ and respectively $\angle N O C$.

a) If $\mathrm{m}(\angle \mathrm{AOD})=82^{\circ}$, determine the measures of angles $\angle M O B, \angle N O C$ and $\angle N O P$.

b) If $\mathrm{m}(\angle \mathrm{MOP})=139^{\circ}$, determine the measure of angle $\angle$ AOD.
Solution and grading criteria:

a)

![](https://cdn.mathpix.com/cropped/2024_06_07_66b8edc299071187ed96g-3.jpg?height=775&width=1005&top_left_y=846&top_left_x=343)

$$
\begin{aligned}
& {\left[OM \text { bisector of } \angle AOD \Rightarrow \mathrm{m}(\angle MOD) = \mathrm{m}(\angle MOA) = 41^{\circ} \text { and }} \\
& \mathrm{m}(\angle MOB) = 180^{\circ} - \mathrm{m}(\angle AOM) = 139^{\circ}
\end{aligned}
$$

b)

Let $\mathrm{m}(\angle AOD) = x$

$\mathrm{m}(\angle MON) = \mathrm{m}(\angle NOB) = 90^{\circ} - \frac{x}{4}$.

$\mathrm{m}(\angle NOP) = 45^{\circ} + \frac{3x}{8}$

$1 \text{p}$

$\mathrm{m}(\angle MOP) = \mathrm{m}(\angle NOP) + \mathrm{m}(\angle MON)$

$45^{\circ} + \frac{3x}{8} + 90^{\circ} - \frac{x}{4} = 139^{\circ}$.

$1 \text{p}$

$x = 32^{\circ}$, thus $\mathrm{m}(\angle AOD) = 32^{\circ}$.