PROBLEM 3. Consider the acute angle $\widehat{X O Y}$. In the half-plane determined by $O X$ and not containing the ray $[O Y$, the rays $[O A$ and $[O B$ are drawn, perpendicular to $[O X$ and $[O Y$ respectively. The ray $[O C$ is the bisector of the angle $\widehat{B O X}$.

a) If the measure of the angle $\widehat{A O C}$ is $16^{\circ}$ greater than the measure of the angle $\widehat{X O Y}$, determine $m(\widehat{X O Y})$;

b) Show that if $[O B$ is the bisector of $\widehat{A O C}$, then $[O X$ is the bisector of $\widehat{C O Y}$.
PROBLEM 3. Consider the acute angle $\widehat{X O Y}$. In the half-plane determined by $O X$ and in which the ray $[O Y$ is not located, the rays $[O A$ and $[O B$ are drawn, perpendicular to $[O X$ and respectively, to $[O Y$. The ray $[O C$ is the bisector of the angle $\widehat{B O X}$.

a) If the measure of the angle $\widehat{A O C}$ is $16^{\circ}$ greater than the measure of the angle $\widehat{X O Y}$, determine $m(\widehat{X O Y})$;

b) Show that if $[O B$ is the bisector of $\widehat{A O C}$, then $[O X$ is the bisector of $\widehat{C O Y}$.

## Grading Rubric.

(1p) a) From $m(\widehat{A O B})=90^{\circ}-m(\widehat{X O B})=m(\widehat{X O Y})$

(1p) and $m(\widehat{A O C})=m(\widehat{A O B})+m(\widehat{B O C})=m(\widehat{X O Y})+m(\widehat{B O C})$

(1p) we obtain that $m(\widehat{B O C})=m(\widehat{A O C})-m(\widehat{X O Y})=16^{\circ}$ and $m(\widehat{X O B})=2 \cdot m(\widehat{B O C})=32^{\circ}$.

(1p) Therefore, $m(\widehat{X O Y})=90^{\circ}-m(\widehat{X O B})=90^{\circ}-32^{\circ}=58^{\circ}$.

$(3 p)$

b) Since $m(\widehat{A O B})=m(\widehat{B O C}), m(\widehat{B O C})=m(\widehat{X O C})$ and $m(\widehat{A O B})=m(\widehat{X O Y})$, it follows that $m(\widehat{X O C})=m(\widehat{X O Y})$.