(本题3分)(2022$cdot$浙江$cdot$瑞安市安阳镇滨江中学三模)如图,已知$triangleABCD$的面积为24,以$B$为位似中心,作$triangleABCD$的位似图形$triangleEBFG$,位似图形与原图形的位似比为$frac{2}{3}$,连接$AG、DG$.则$triangleADG$的面积为$qquad$.<ImageHere>
【分析】延长$EG$交$CD$于点$H$,由题意可得四边形$AEHD$是平行四边形,则可得此平行四边形的面积为8,从而可得$triangleADG$的面积.【详解】延长$EG$交$CD$于点$H$,如图,$because$四边形$ABCD$是平行四边形,四边形$EBFG$是平行四边形,$thereforeAB//CD,AD//BC;BF//EG$,$thereforeAD//EG$,$therefore$四边形$AEHD$是平行四边形,$thereforeS_{triangleADG}=frac{1}{2}S_{squareAEHD}$$because$位似图形与原图形的位似比为$frac{2}{3}$,$thereforeBE=frac{2}{3}AB$,即$AE=frac{1}{3}AB$,$thereforeS_{squareAEHD}=frac{1}{3}S_{squareABCD}=8$,<ImageHere><ImageHere>故答案为:4.【点睛】本题考查了位似图形的性质,平行四边形的性质与判定,掌握这些性质是解题的关键.