Question

.如图2-13,已知⊙O外一点P,作⊙O的切线PQ,Q为切点,过PQ中点M,作割线MAB,交⊙O于点A、B,连结PA并延长交⊙O于C,连结PB交⊙O于点D,求证:CD∥PQ.

图2-13

Answer 1

思路分析:本题要证CD∥PQ,只要证∠ACD ＝∠APQ，又∠ACD＝∠ABD,因而只需证∠ABD =∠APQ,这可利用相似三角形证得.

证明:∵PQ切⊙O于Q,?

∴MQ²=MA·MB.?

又∵M为PQ中点,即MQ =MP,?

∴MP²=MA·MB,即=.?

又∠AMP =∠PMB,?

∴△AMP∽△PMB.∴∠MPA =∠ABD.?

又∵∠ABD =∠ACD,?

∴∠MPA =∠ACD.∴CD∥PQ.