2018 USAMO 第 5 题

In convex cyclic quadrilateral $ABCD,$ we know that lines $AC$ and $BD$ intersect at $E,$ lines $AB$ and $CD$ intersect at $F,$ and lines $BC$ and $DA$ intersect at $G.$ Suppose that the circumcircle of $\triangle ABE$ intersects line $CB$ at $B$ and $P$, and the circumcircle of $\triangle ADE$ intersects line $CD$ at $D$ and $Q$, where $C,B,P,G$ and $C,Q,D,F$ are collinear in that order. Prove that if lines $FP$ and $GQ$ intersect at $M$, then $\angle MAC = 90^{\circ}.$

在凸循环四边形$ABCD,$中，我们知道$AC$和$BD$相交于$E，$AB$和$CD$相交于$F,$，$BC$和$DA$相交于$G。$假设$\三角形ABE$的外接圆与$CB$相交于$B$和$P$，$\三角形ADE$的外接圆与$CD$相交于$\triangle ADE$$D$和$Q$，其中$C、B、P、G$和$C、Q、D、F$按此顺序共线。证明如果线$FP$和$GQ$相交于$M$，则$\angle MAC = 90^{\circ}.$