如图,△ABC内接于⊙O,且AB=AC,D是AC上一点,AD与BC交于E,AF⊥DB,垂足为F.(1)求证:∠ADB=∠CDE;

2025-05-18 11:51:05
推荐回答(1个)
回答1:

(1)证明:∵AB=AC,
∴∠ABC=∠BCA=∠ADB,
∵四边形ABCD是圆内接四边形,
∴∠CDE=∠ABC,
∴∠ADB=∠CDE;

(2)解:作AM⊥CD于点M,
∵AB=10,AF=6,
∴BF=8,
∵AD平分∠BDM,AM=AF=6,
∴△ACM≌△ABF,
∴CM=BF=8,
∴DF=DM=CM-CD=2.
∴BD=BF+DF=10=AB.
∴∠BAD=∠ADB=∠ADM,
∴S△DBC=S△ADC=

1
2
CD×AM=18.