题目内容
(本小题满分10分)选修4-1:几何证明选讲
如图,AB是圆O的直径,C是半径OB的中点,D是OB延长线上一点,且BD=OB,直线MD与圆O相交于点M、T(不与A、B重合),DN与圆O相切于点N,连结MC,MB,OT.
(1)求证:
;
(2)若
,试求
的大小.
如图,AB是圆O的直径,C是半径OB的中点,D是OB延长线上一点,且BD=OB,直线MD与圆O相交于点M、T(不与A、B重合),DN与圆O相切于点N,连结MC,MB,OT.
(1)求证:

(2)若



(1)证明见解析
(2)
(2)

(1)证明:因MD与圆O相交于点T,由切割线定
理
,
,得
,设半径OB=
,
因BD=OB,且BC=OC=
,
则
,
,
所以
(2)由(1)可知,
,
且
,
故
∽
,所以
;
根据圆周角定理得,
,则
理




因BD=OB,且BC=OC=

则


所以

(2)由(1)可知,

且

故



根据圆周角定理得,



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