题目内容
如图,BD为⊙O的直径,点A是弧BC的中点,AD交BC于E点,AE=2,ED="4. "
(1)求证:
~
;
(2) 求
的值;
(3)延长BC至F,连接FD,使
的面积等于
,求
的度数.

(1)求证:


(2) 求

(3)延长BC至F,连接FD,使



(1)证明略
(2)

(3)60°
(1)
证明:∵点A是弧BC的中点,
∴∠ABC=∠ADB.
又∵∠BAE=∠DAB,
∴△ABE∽△ADB.…………………………………………………2分
(2)解
∵△ABE∽△ADB,
∴AB2=2×6=12.
∴AB=2
.
在Rt△ADB中,tan∠ADB=
………………………4分
(3)解:连接CD,
∵tan∠ADB=
,∴∠ADB=30°.
又∵A为
的中点,∴∠ABC=∠ADB=30°.
∵∠A=90°,∠ABD=60°.
∴∠DBC=30°.
∴CD=AB=2
,BE=DE=4.
又∵S△BDF=8
,
∴BF=8.
∴EF=4.
又∵∠FED=∠EBD+∠EDB=60°,
∴△EFD为等边三角形.
∴∠EDF=60°…………………………………………………………7分
证明:∵点A是弧BC的中点,
∴∠ABC=∠ADB.
又∵∠BAE=∠DAB,
∴△ABE∽△ADB.…………………………………………………2分
(2)解
∵△ABE∽△ADB,
∴AB2=2×6=12.
∴AB=2

在Rt△ADB中,tan∠ADB=

(3)解:连接CD,
∵tan∠ADB=

又∵A为

∵∠A=90°,∠ABD=60°.
∴∠DBC=30°.
∴CD=AB=2

又∵S△BDF=8

∴BF=8.
∴EF=4.
又∵∠FED=∠EBD+∠EDB=60°,
∴△EFD为等边三角形.
∴∠EDF=60°…………………………………………………………7分

练习册系列答案
相关题目