提问[2013年高考数学江苏卷第16题第(1)问]如图1所示,在三棱锥S-ABC中,平面SAB⊥平面SBC,AB⊥BC,AS=AB,过A作AF⊥SB,垂足为F,点E,G分别是棱SA,SC的中点.求证:平面EFG∥平面ABC.对上面这道题,我是这样证的:因为点E,G分别是棱SA,SC的中点,所以EG是△SAC的中位线,故EG∥AC.由AS=AB,AF⊥SB可得F是等腰三角形SAB底边SB的中点,而点 Asking questions [2013 college mathematics Jiangsu volume 16 questions (1) asked] As shown in Figure 1, in the triangular pyramid S-ABC, the plane SAB⊥ plane SBC, AB ⊥ BC, AS = AB, A for AF ⊥ SB, vertical foot for the F, points E, G, respectively, the edge SA, SC of the middle. Prove: plane EFG ∥ plane ABC. On the above question, I hereby certify: because point E, G are SA , So EG is the median line of △ SAC, so EG∥AC.From AS = AB, AF⊥SB, F is the midpoint of the bottom SB of the isosceles triangle SAB, and the point