试题分析:(1)易证△ADE≌△CDG,即可得出AE:CG=1;
(2)与(1)类似,证明△ADE≌△CDG,即可得出AE:CG=1;
(3)证明△ADE∽△CDG即可.
试题解析:(1)∵正方形ABCD和正方形DEFG,∴AD=CD,DE=DG,∠ADC=∠EDG=90°,∴∠ADE=∠CDG,在△AED和△CGD中,∵AD=CD,∠ADE=∠CDG ,DE=DG,∴△ADE≌△CDG,∴AE=CG,∴AE:CG=1;
(2)成立,理由如下:
∵正方形ABCD和正方形DEFG,∴AD=CD,DE=DG,∠ADC=∠EDG=90°,∴∠ADE=∠CDG,在△AED和△CGD中,∵AD=CD,∠ADE=∠CDG ,DE=DG,∴△ADE≌△CDG,∴AE=CG,∴AE:CG=1;
(3)∵矩形ABCD和矩形DEFG,∴∠ADC=∠EDG=90°,∴∠ADE=∠CDG,∵
,
,∴
,∴△ADE∽△CDG,∴AE:CG=AD:DC=4:6=2:3.