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九年级(上)期末数学试卷集(328题)


如图,BDO的直径,AB=ACADBC于点EAE=2ED=4

1)求证:ABE∽△ADB,并阴影部分面积


2)延长DBF,使得BF=BO,连接FA,试判断直线FAO的位置关系,并说明理由. 




知识点:试卷11


参考答案:见解析


解析:

17分)证明:AB=AC∴∠ABC=C

∵∠C=D∴∠ABC=D


∵∠BAE=EAB


∴△ABE∽△ADB...........................................................2



AB2=ADAE=AE+EDAE=2+4×2=12


AB=....................................................................3


BDO的直径,∴∠BAD=90°



∴∠ABE=30°..................................................................5


∴∠ADB=30°


∴∠CBD=30°∴AC//BD.......................................6


连接OAOBOA=,∠AOC=60°


........................7


24分)直线FAO相切,.........................................................1


理由如下:


∵∠ABC=30°,AB=AC


∴∠C=30°∴∠AOB=60°


.是等边三角形..............................................................2


AB=BO,∠BAO=∠ABO=60°


∴∠FAB=∠ABO=30°∴∠OAF=90°........................................3


∴直线FAO相切...............................................................................4

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