题目内容
已知AB是⊙
的直径,弦AC平分
,AD
CD于D,BE
CD于E。
求证:⑴CD是⊙
的切线;
⑵





求证:⑴CD是⊙

⑵


⑴连结OC …………………1′

∴ ∠OAC=∠OCA
∵ AC平分∠BAC
∴ ∠DAC=∠OAC
∴ ∠OCA=∠DAC …………………2′
∴ AD∥OC
∵ AD⊥CD
∴ OC⊥CD …………………3′
∴ CD是⊙的切线 …………………4′
⑵ 连结BC,延长AC交BE的延长线于M …………………5′
∵ AD⊥DE BE⊥DE
∴ AD∥BE
∴ ∠M=∠DAC
∵ ∠DAC=∠BAM
∴ ∠BAM=∠M
∴ BA="BM " …………………6′
∵ AB是直径
∴ ∠ACB=90
∴ AC=MC
又 ∵ ∠M=∠DAC ∠D=∠CEM AC=MC
∴
∴ DC="EC " …………………7′
(若用平行线分线段成比例定理证明,正确得分)
∴ ∠DAC=∠BCE ∠ADC=∠CEB
∴
ADC~
CEB …………………8′
∴

∴
∴
…………………9′
说明:本题还有其它证法,若正确合理得分。

∴ ∠OAC=∠OCA
∵ AC平分∠BAC
∴ ∠DAC=∠OAC
∴ ∠OCA=∠DAC …………………2′
∴ AD∥OC
∵ AD⊥CD
∴ OC⊥CD …………………3′
∴ CD是⊙的切线 …………………4′
⑵ 连结BC,延长AC交BE的延长线于M …………………5′
∵ AD⊥DE BE⊥DE
∴ AD∥BE
∴ ∠M=∠DAC
∵ ∠DAC=∠BAM
∴ ∠BAM=∠M
∴ BA="BM " …………………6′
∵ AB是直径
∴ ∠ACB=90

∴ AC=MC
又 ∵ ∠M=∠DAC ∠D=∠CEM AC=MC
∴

∴ DC="EC " …………………7′
(若用平行线分线段成比例定理证明,正确得分)
∴ ∠DAC=∠BCE ∠ADC=∠CEB
∴


∴


∴

∴

说明:本题还有其它证法,若正确合理得分。
略

练习册系列答案
相关题目