【题目】(2017山东省莱芜市)如图,在矩形ABCD中,BEAC分别交ACAD于点FE,若AD=1,AB=CF,则AE=______


参考答案:

【答案】

【解析】解:四边形ABCD是矩形,BC=AD=1,∠BAF=∠ABC=90°,∴∠ABE+∠CBF=90°,∵BEAC,∴∠BFC=90°,∴∠BCF+∠CBF=90°,∴∠ABE=∠FCB,在ABEFCB中,∵∠EAB=∠BFC=90°,AB=CF,∠ABE=∠FCB,∴△ABE≌△FCB,∴BF=AEBE=BC=1,∵BEAC,∴∠BAF+∠ABF=90°,∵∠ABF+∠AEB=90°,∴∠BAF=∠AEB,∵∠BAE=∠AFB,∴△ABE∽△FBA,∴,∴,∴AE=BF=AB2,在Rt△ABE中,BE=1,根据勾股定理得,AB2+AE2=BE2=1,∴AE+AE2=1,∵AE>0,∴AE=故答案为:

关闭