EFGH是矩形ABCD的内接矩形,且EF:FG=3:1,AB:BC=2:1,则AH:AE=_.

问题描述:

EFGH是矩形ABCD的内接矩形,且EF:FG=3:1,AB:BC=2:1,则AH:AE=______.

设AH=a,AE=b,∵∠CFG=∠DGH=∠AHE=∠FEB,∠HAE=∠FCG=90°,AE=CG,∴△AHE≌△CFG(AAS);由三角分别相等可判定△CGF∽△BFE,∴EBFC=EFFG=FBGC=3,即EB=3a,BF=3b,∵AB:BC=2:1,∴3a+b3b+a=2,即a=5b,故AH...