9. $A B C D$ is a rectangle whose diagonals intersect at point $O . E$ is a point on $A B$ such that $C E$ bisects $\angle B C D$. If $\angle A C E=15^{\circ}$, find $\angle B O E$.
(A) $60^{\circ}$
(B) $65^{\circ}$
(C) $70^{\circ}$
(D) $75^{\circ}$
(E) $80^{\circ}$
9. Answer: (D)
Since $C E$ bisects $\angle B C D, \angle B C E=45^{\circ}$. Thus $\angle C E B=45^{\circ}$ also and $\triangle C B E$ is isosceles, Therefore $B C=B E$.
Now $\angle B C O=45^{\circ}+15^{\circ}=60^{\circ}$. As $C O=B O$, we conclude that $\triangle C O B$ is equilateral. Thus $B C=B O=B E$ giving us an isosceles triangle $O B E$. Since $\angle O B E=30^{\circ}$, thus $\angle B O E=75^{\circ}$.