1. As shown in the figure: In $\triangle A B C$, the angle bisector of $\angle A$ intersects $B C$ at $D$, and intersects the circumcircle of $\triangle A B C$ at $P$. A circle is drawn with $A D$ as a chord, intersecting $A B$ at $M$ and the extension of $A C$ at $N$. $P M$ intersects $B C$ at $G$. If the area of $\triangle B M G$ is $S_{1}$, and the area of quadrilateral $P N C G$ is $S_{2}$, compare the sizes of $S_{1}$ and $S_{2}$.
1. $S_{1}=S_{2}$. Proof as follows:

Extend $B C$ to intersect the circle with $A D$ as the chord at $L$.
Connect $L N$, $L P$, $L M$, $B P$, $C P$.
Then $\angle D L N=\angle D A N=\angle P A C=\angle P A B=\angle D C P$.
$\therefore P C \parallel L N$.
Let $L P$ intersect $C N$ at $K$,
then $S_{\triangle P K N}=S_{\triangle O K L}$.
$\therefore \quad S_{\text{quadrilateral } P N G G}=S_{\triangle R G L}$.
Also, $\angle M L D=\angle M A D=\angle P A C=\angle P B C$,
$\therefore B P \parallel M L$.
$\therefore S_{\triangle M D}=S_{\triangle C P L L}$.
$\therefore S_{\triangle M A K C}=S_{\text{quadrilateral } P N N C}$, i.e., $S_{1}=S_{2}$. Proof completed.