题目内容

如图2-1-21,已知AD为锐角△ABC的外接圆O的直径,AEBCE,交外接圆于F,

图2-1-21

(1)求证:∠1=∠2;

(2)求证:AB·AC=AE·AD;

(3)作OHAB,垂足为H.求证:.

思路分析:(1)∠1与∠2均为圆周角,要证它们相等,只需证所对的弧相等,弧BD与弧FC夹在BCDF之间,只需证DFBC即可.?

(2)要证等积式,可先证比例式=,而这可由△ABD∽△AEC证得.?

(3)要证,联想到中位线定理,可先证.

证明:(1)连结DF,∵AD为直径,∴∠AFD =90°.?

BCAF,∴DFBC.?

=.∴∠1=∠2.?

(2)连结BD,∵AD为直径,∴∠ABD =90°.?

AEBC,∴∠AEC=90°.?

∴∠ABD =∠AEC.?

又∠1=∠2,?

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

=,?

AB·AC =AE·AD.?

(3)连结CF,∵AD为直径,∴∠ABD =90°.?

OHAB,∴OHBD.?

HAB中点,即OH为△ABD的中位线.?

.?

=,∴BD =CF.?

.

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