Question

如圖2-1-21,已知AD為銳角△ABC的外接圓O的直徑,AE⊥BC于E,交外接圓于F,

圖2-1-21

(1)求證:∠1=∠2;

(2)求證:AB·AC=AE·AD;

(3)作OH⊥AB,垂足為H.求證:.

Answer 1

思路分析:(1)∠1與∠2均為圓周角,要證它們相等,只需證所對(duì)的弧相等,弧BD與弧FC夾在BC與DF之間,只需證DF∥BC即可.?

(2)要證等積式,可先證比例式=,而這可由△ABD∽△AEC證得.?

(3)要證,聯(lián)想到中位線定理,可先證.

證明:(1)連結(jié)DF,∵AD為直徑,∴∠AFD =90°.?

又BC⊥AF,∴DF∥BC.?

∴ =.∴∠1=∠2.?

(2)連結(jié)BD,∵AD為直徑,∴∠ABD =90°.?

又AE⊥BC,∴∠AEC=90°.?

∴∠ABD =∠AEC.?

又∠1=∠2,?

∴△ABD∽△AEC(或由∠1=∠2,∠ACB =∠ADB可知△ABD∽△AEC).?

∴=,?

即AB·AC =AE·AD.?

(3)連結(jié)CF,∵AD為直徑,∴∠ABD =90°.?

又OH⊥AB,∴OH∥BD.?

∴H為AB中點(diǎn),即OH為△ABD的中位線.?

∴.?

又 =,∴BD =CF.?

∴.