:如图,在Rt△ABC中,∠ABC = 90°,半圆O切BC于点B,切AC于点D,交AB于点E,BC= BE =2,求AE和AD的长

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:如图,在Rt△ABC中,∠ABC = 90°,半圆O切BC于点B,切AC于点D,交AB于点E,BC= BE =2,求AE和AD的长

:如图,在Rt△ABC中,∠ABC = 90°,半圆O切BC于点B,切AC于点D,交AB于点E,BC= BE =2,求AE和AD的长
:如图,在Rt△ABC中,∠ABC = 90°,半圆O切BC于点B,切AC于点D,交AB于点E,BC= BE =2,求AE和AD的长

:如图,在Rt△ABC中,∠ABC = 90°,半圆O切BC于点B,切AC于点D,交AB于点E,BC= BE =2,求AE和AD的长
已知,CB和CD和圆O分别相切于点B、D,可得:CD = BC = 2 .
设 AE=x ,AD = y ,则 OA = 1+x ,AC = 2+y .
OD/OA = sin∠A = BC/AC ,即有:1/(1+x) = 2/(2+y) ,可得:y = 2x .
在Rt△OAD中,由勾股定理可得:1+4x² = (1+x)² ,解得:x = 2/3 .
可得:AE = 2/3 ,AD = 4/3 .