, 2009

Solution 1. First, we prove the “if” direction by assuming that ray $BG$ bisects $\angle CBD$; that is, we assume that $\widehat{DQ} = \widehat{CQ}$. It is easy to see that $ABCD$ is an isosceles trapezoid with $AD = BC$. In particular, $\widehat{AD} = \widehat{BC}$ and $\widehat{AC} = \widehat{BD}$. Because $ABCPD$ is cyclic, it follows that
$$\angle APC = \frac{\widehat{AC}}{2} = \frac{\widehat{BD}}{2} = \angle BCD = \angle SCD \quad \text{and} \quad \angle APD = \frac{\widehat{AD}}{2} = \frac{\widehat{BC}}{2} = \angle BDC = \angle RDC.$$
Because $RS \parallel DC$, it follows that $180^\circ = \angle GRD + \angle RDC = \angle GRD + \angle APD$ and $180^\circ = \angle GSC + \angle SCD = \angle GSC + \angle APC$; that is, both $GSCP$ and $GRDP$ are cyclic. Hence, $\angle GPR = \angle GDR$ and $\angle GPS = \angle GCS$. In particular, we have
$$\angle RPS = \angle GPR + \angle GPS = \angle GDR + \angle GCS. \qquad (1)$$
Let $K$ be the intersection of segments $BQ$ and $CD$. We have $\angle CBK = \angle QBD$ and $\angle KCB = \angle DCB = \angle DQB$; that is, triangles $CBK$ and $QBD$ are similar to each other. Because $RG \parallel CD$, we have $BG/GK = BR/RD$. This means that $G$ and $R$ are the corresponding points in the similar triangles $CBK$ and $QBD$. Consequently, we have $\angle BCG = \angle BQR$. In exactly the same way, we can show that $\angle BDG = \angle BQS$. Combining the last two equations together with (1) yields
$$\angle RQS = \angle BQS + \angle BQR = \angle BDG + \angle BCG = \angle RDG + \angle SCG = \angle RPS,$$
from which it follows that $PQRS$ is cyclic.

Second, we prove the “only if” direction by assuming that $PQRS$ is cyclic. Let $\gamma$ denote the circumcircle of $PQRS$. We approach indirectly by assuming that ray $BG$ does not bisect $\angle CBD$. Let $G_1$ be the point on segment $RS$ such that ray $BG_1$ bisects $\angle CBD$. Let rays $AG_1$ and $BG_1$ meet $\omega$ again at $P_1$ and $Q_1$ (other than $A$ and $B$). By our proof of the “if” part, $P_1Q_1RS$ is cyclic, and let $\gamma_1$ denote its circumcircle.

Hence lines $RS, PQ, P_1Q_1$ are the radical axes of pairs of circles $\gamma$ and $\gamma_1$, $\gamma$ and $\omega$, $\gamma_1$ and $\omega$, respectively. Because point $Q_1$ is the midpoint of arc $\widehat{CD}$ (not including $A$ and $B$), $P_1Q_1 \parallel CD$, implying that lines $P_1Q_1$ and $RS$ intersect. Let $X$ denote this intersection; then $X$ is the radical center of $\omega, \gamma, \gamma_1$. In particular, line $PQ$ also passes through $X$, giving the following configuration.

There are two possibilities for the position of line $PQ$, namely, (a) both $P$ and $Q$ lie on minor arc $\widehat{P_1Q_1}$; (b) one of $P$ and $Q$ lies on minor arc $\widehat{DQ_1}$ and the other lies on minor arc $\widehat{P_1B}$. If $G$ lies on segment $RG_1$, then $Q$ lies on minor arc $\widehat{DQ}$, and we must have (b). But in this case, $P$ must lie on minor arc $\widehat{Q_1P_1}$, violating (b). If $G$ lies on segment $G_1S$, then $P$ must lie on minor arc $\widehat{P_1B}$, and again we must have (b). But in this case, $Q$ must lie on minor arc $\widehat{Q_1C}$, violating (b). In every case, we have a contradiction. Hence our assumption was wrong, and ray $BG$ bisects $\angle CBD$.

Solution 2. We present another approach for the “if” direction.

Let rays $CG$ and $DG$ meet $\omega$ again at $E$ and $F$, respectively. Let $R_1$ denote the intersection of segments $BD$ and $QE$, and let $S_1$ denote the intersection of segments $BC$ and $QF$. Applying Pascal's theorem to cyclic hexagon $BDFQEC$ shows that $R_1, G, S_1$ are collinear. Because
$$\angle R_1EG = \angle QEC = \frac{\widehat{CQ}}{2} = \frac{\widehat{DQ}}{2} = \angle DBQ = \angle R_1BG,$$
we deduce that $EBGR_1$ is cyclic. Because $EBGR_1$ and $EBCD$ are cyclic, we have
$$\angle BR_1S_1 = \angle BR_1G = \angle BEG = \angle BEC = \angle BDC,$$
from which it follows that $R_1S_1 \parallel CD$; that is, $R_1 = R$ and $S_1 = S$. Therefore, (1) becomes
$$\angle RPS = \angle GDR + \angle GCS = \angle FDB + \angle BCE = \angle FQB + \angle BQE = \angle FQE = \angle RQS,$$
implying that $PQRS$ is cyclic.