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已知在Rt△ABC中,∠ACB=90°,CD⊥AB于D,则下列各式中成立的有:
A) AC^2:BC^2 = AD:BD
B) AC^2:BD^2 = AC:BC
C) AC:BD = AD:BD
D) AC:CD = CD:BD
人气:408 ℃ 时间:2020-06-30 22:14:06
解答
选A
因为△ABC与△CBD、△ACD相似
AC:AB=AD:AC,AC^2=AB*AD
BC:BD=AB:BC,BC^2=AB*BD
所以AC^2:BC^2=(AB*AD):(AB*BD)
=AD:BD
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