In the figure to the right, let $$\mathrm{OA} = 6, \quad \mathrm{OB} = 3, \quad \angle\mathrm{AOB} = 120^\circ,$$ and let the point Q denote the point of intersection of the bisector of $\angle\mathrm{XAB}$ and the bisector of $\angle\mathrm{ABY}$. Let P denote the point of intersection of segment AB and segment OQ. We are to find the length of segment PQ.
(1) First of all, we see that $\mathrm{AB} = \square\sqrt{\mathbf{A}}$ and that the area of triangle OAB is $\mathbf{BC}\sqrt{\mathbf{D}}$.
(2) For $\mathbf{F}$ and $\mathbf{G}$ in the following, choose the correct answer from among choices (0) $\sim$ (4), just below. (0) AB (1) AP (2) AQ (3) BP (4) BQ
Since AQ is the bisector of the exterior angle of $\angle\mathrm{A}$ of triangle OAP and BQ is the bisector of the exterior angle of $\angle\mathrm{B}$ of triangle OBP, we have $$\begin{aligned} \mathrm{OQ} : \mathrm{PQ} &= \mathrm{OA} : \mathbf{F} \\ &= \mathrm{OB} : \mathbf{G} \end{aligned}$$
(3) Thus we see $\mathrm{AP} = \mathbf{H}\sqrt{\mathbf{I}}$. Since $\angle\mathrm{AOP} = \mathbf{JK}^\circ$, we have $\mathrm{OP} = \square\mathbf{L}$. Hence we have $\mathrm{PQ} = \mathbf{M} + \mathbf{N}\sqrt{\mathbf{O}}$.